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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1101. |
In the circuit element given here, if the potential at point `B = V_(B) = 0`, then the potentials of `A` and `D` are given as A. `V_(A)=-1.5V,V_(D)=+2V`B. `V_(A)=+1.5V,V_(D)=+2V`C. `V_(A)=+1.5V,V_(D)=+0.5V`D. `V_(A)=+1.5V,V_(D)=-0.5V` |
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Answer» Correct Answer - 4 `V_(B)=0` `V_(A)-V_(B )`=(current)`xx` (Resistance of AB) `V_(A)-0=1xx 1.5` `therefore V_(A)=1.5` `V_(B )-V_(C)=` (current) `xx` (resistance ofBC) `0-V_(C)=1xx2.5` or `V_(C)=-2.5` volt For cell `V_(D)-V_(C)=2` volt `therefore V_(D)-(-2.5)=2` or `V_(D)+2.5=2` or `V_(D)=2-2.5=-0.5` `therefore V_(D)=-0.5` volt `therefore V_(A)=+1.5,VD=-0.5V` |
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| 1102. |
In the figure shown, the total resistance between `A` and `B` is A. `12 Omega`B. `4 Omega`C. `6 Omega`D. `8 Omega` |
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Answer» Correct Answer - D (d) The last two resistance are out of cicuit. Now `8 Omega` is in parallel with `(1 + 1 + 4 + 1) Omega`. `:. R = 8 Omega || 8 Omega = (8)/(2) = 4 Omega R_(AB) = 4 + 2 + 2 = 8 Omega` |
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| 1103. |
Find the equivalent resistance of the circuit between points `A` and `B` shown in figure is (: each brach if of resistance `0 1 Omega`) A. `(22)/(25) Omega`B. `(12)/(25) Omega`C. `(22)/(35) Omega`D. `(12)/(35) Omega` |
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Answer» Correct Answer - C (c ) By perpendicular Axis symmetry all point `1, 2, 3,` are at same potential therefore junction on this line can be redrawn as `R_(AB) = (22)/(35) R` |
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| 1104. |
What is the equivalent resistance between the point `A` and `B` of the network ? A. `(57)/(7)Omega`B. `8Omega`C. `6Omega`D. `(57)/(5)Omega` |
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Answer» Correct Answer - B |
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| 1105. |
The effective resistance between points P and Q of the electrical circuit shown in the figure is (a) `(2Rr)/(R+r)` (b)`(8R(R+r))/(3R +r)` (c) `2r + 4R` (d) `(5R)/2 + 2r` . .A. `2R//(R +r)`B. `8R(R +r)//(3R+r)`C. `2r +4R`D. `5R//2 +2r` |
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Answer» Correct Answer - A |
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| 1106. |
The equivalent resistance between the point `P` and `Q` in the network given here is equal to (given `r = (3)/(2) Omega)` A. `(1)/(2)Omega`B. `1 Omega`C. `(3)/(2)Omega`D. `2 Omega` |
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Answer» Correct Answer - B |
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| 1107. |
In the circuit element given here, if the potential at point `B = V_(B) = 0`, then the potentials of `A` and `D` are given as A. `V_(A) =- 1.5, V_(D) = +2V`B. `V_(A) = +1.5V,V_D) =+2V`C. `V_(A) = +1.5V,V_(D) = +0.5V`D. `V_(A) = +1.5V,V_(D) =- 0.5V` |
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Answer» Correct Answer - D |
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| 1108. |
On a pT diagram, a cyclic process is performed as shown. Where is the volume maximum? A. aB. bC. cD. d |
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Answer» Correct Answer - D |
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| 1109. |
Draw a colour code for 42k `Omega+-10%` carbon resistance. |
| Answer» According to colour code colour for digit 4 is yellow, for digit 2 it is red, for 3 colour is orange and 10% tolerance is represented b y silver colour. So colour code should be yellow, red,orange and sivler. | |
| 1110. |
For the given part of circuit calculate potential `V_(p)` of centre point P and then find current `I_(1)`. |
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Answer» Let potential of point P is `V_(P)`. By using KCL at point `PI_(1)+I_(2)+I_(3)=0` so `(V_(p)-6)/(20)+(V_(p)-10)/(10)+(V_(p)-5)/(30)=0` `rArr 3(V_(p)-6)+6(V_(p)-10)+2(V_(p)-5)=0rArr11V_(p)-88=0rArrV_(p)=8` volt an d current `I_(1)=(8-6)/(20)=0.1A` |
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| 1111. |
What is resistance of following resistor. |
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Answer» Number for yellow is 4, Number of violet is 7 Brown colour gives multiplier `10^(1)` , Gold gives a tolerance of `+-5%` So resistance of resistor is `47xx10^(1)Omega+-5%=470+-5%Omega`. |
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| 1112. |
Find the ratio ofcurrents as measured by ammeter in two cases when the key is open and when the key is closed : A. `9//8`B. `10//11`C. `8//9`D. None of the above |
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Answer» Correct Answer - D |
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| 1113. |
A wire of length 100 cm is connected to a cell of emf 2 V and negligible internal resistance. The resistance of the wire is `3 Omega`. The additional resistance required to produce a potential drop of 1 milli volt per cm isA. `60Omega`B. `47Omega`C. `57Omega`D. `35Omega` |
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Answer» Correct Answer - C |
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| 1114. |
The reading of the ammeter in the following figure will be A. `0.8` AB. `0.6` AC. `0.4` AD. `0.2` A |
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Answer» Correct Answer - C |
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| 1115. |
If power dissipated in the `9 Omega` resistor in the resistor shown is `36 W`, the potential difference across the `2 Omega` resistor is A. 8 VB. 10 VC. 2 VD. 4 V |
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Answer» Correct Answer - B |
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| 1116. |
In the given circuit diagram, potential difference across `2 Omega` resistance is A. `4 V`B. `6 V`C. `20 V`D. zero |
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Answer» Correct Answer - D (d) Since it is a balanced Wheatstone bridge, potential difference across `2 Omega` resistance is zero. |
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| 1117. |
In a metere bridge, the wire consists of two parts one of length 30 cm and of radius r and the other of radius 2r. Where will the null point occur if the resistance in the left and right gaps are `5Omega` and `8Omega` respectively? The material of the wires is the same. |
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Answer» Correct Answer - 18.3 cm We have `("left gap resistance")/("right gap resistance")=(5)/(8)=0.625` `("left length resistance")/("right length resistance")=((30)/(r^(2)))/((70)/(4r^(2)))=1.7` Hence the null point is on the thin wire. Let it be at a distance x. Then `(5)/(8)=(rhoxx(x))/((pir^(2))/(rhoxx((30-x))/(pir^(2))+(rho70)/(pi4r^(2))))=(4x)/(190-4x)` or `x=18.3cm` |
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| 1118. |
When an ideal voltmeter is connected between the [points E and F the reading of the meter is `V_(0)`. When an ideal ammeter is connected between E and F, reading is `I_(0)` find the current I through resistor R connected between E and F. |
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Answer» When an ammeter is put between E and F `I_(0)=(V_(0))/(r+0)(thereforeR_(A)=0)impliesr=(V_(0))/(I_(0))` When a resistance R is put between E and F `I=(V_(0))/(R+r)=(V_(0))/(R(V_(0))/(I_(0)))=(V_(0)I_(0))/(V_(0)+I_(0)R)` |
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| 1119. |
If the current in a potentiometer increases, the position of the null point willA. be obtained at a larger length than the previous oneB. be equal to the previous lengthC. be obtained at a smaller length than the previousD. none of these |
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Answer» Correct Answer - C `x=(iR)/LrArr ` as current increases potential gradient increases so balance length decreases. |
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| 1120. |
When a resistance of 2 ohm is connected across terminals of a cell, the current is `0.5 A`. When the resistance is increased to 5 ohm, the current is `0.25 A` The e.m.f. of the cell isA. `1.0 V`B. `1.5 V`C. `2.0 V`D. `2.5 V` |
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Answer» Correct Answer - B (b) Since `i= ((E)/(R +r))`, we get `0.5 = (E)/(2 + r)` `0.25 = (E)/(5 + r)` Dividing (i) by (ii), we get `2 = (5 + r)/(2 + r) implies r = Omega` `:. 0.5 = (E)/(2 + 1) implies E = 1.5 V` |
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| 1121. |
In the circuit shown, A and V are ideal ammeter and voltmeter respectively. Reading of the voltmeter will be A. 2VB. 1VC. 0.5VD. Zero |
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Answer» Correct Answer - D |
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| 1122. |
When a resistance of 2 ohm is connected across terminals of a cell, the current is `0.5 A`. When the resistance is increased to 5 ohm, the current is `0.25 A` The e.m.f. of the cell isA. `0.5` ohmB. `1.0` ohmC. `1.5` ohmD. `2.0` ohm |
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Answer» Correct Answer - B |
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| 1123. |
The terminal potential difference of a cell when short-circuited is (E = E.M.F. of the cell)A. EB. `E//2`C. ZeroD. `E//3` |
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Answer» Correct Answer - C |
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| 1124. |
A primary cell has an e.m.f. of 1.5 volts, when short-circuited it gives a current of 3 amperes. The internal resistance of the cell isA. 4.5 ohmB. 2 ohmC. 0.5 ohmD. `1//4.5` ohm |
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Answer» Correct Answer - C |
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| 1125. |
The potential difference in open circuit for a cell is 2.2 volts. When a 4 ohm resistor is connected between its two electrodes the potential difference becomes 2 volts. The internal resistance of the cell will beA. 1ohmB. 0.2 ohmC. 2.5 ohmD. 0.4 ohm |
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Answer» Correct Answer - D |
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| 1126. |
Two cells, having the same emf, are connected in series through an external resistance `R`. Cells have internal resistance `r_(1)` and `r_(2) (r_(1) gt r_(2))` respectively. When the circuit is closed, the potentail difference across the first cell is zero the value of `R` isA. `r_(2)-r_(1)`B. `r_(1)-r_(2)`C. `r_(1)+r_(2)`D. `(r_(1)-r_(2))/(2)` |
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Answer» Correct Answer - B `i=(E_(1)+E_(2))/(R+r_(1)+r_(2)),V_(1)=E_(1)-ir_(1)` |
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| 1127. |
A `50V` battery is connected across a 10 ohm resistor. The current is `4.5` amperes. The internal resistance of the battery isA. ZeroB. `0.5` ohmC. `1.1` ohmD. `5.0` ohm |
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Answer» Correct Answer - C |
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| 1128. |
The effective resistance between A and B in the given circuit isA. `20Omega`B. `7Omega`C. `3Omega`D. `6Omega` |
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Answer» Correct Answer - D combination of resistors |
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| 1129. |
When two identical cells are connected either in series or in parallel across a 4 ohm resistor, they send the same current through it. The internal resistance of the cell in ohm isA. `4Omega`B. `2Omega`C. `1Omega`D. `7Omega` |
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Answer» Correct Answer - A when `i_(s)=i_(p),r=R` |
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| 1130. |
How many cells each marked `(6V-12A)` should be connected in mixed grouping so that it may be marked `(24V-24VA)`A. 4B. 8C. 12D. 6 |
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Answer» Correct Answer - A number of rows `=("required current")/("given current")` `=(24A)/(12A)=2=m` Number of cells in each now `=("required potential")/("given potential")` `=(24)/(6)=4=n` `therefore` Total no of cells `=nxxm` `=2xx4=8` |
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| 1131. |
Under what conditions is the current through the mixed grouping of cells maximum ? |
| Answer» The current through the mixed grouping of cells maximum, when (1) Effective emf of all the cells is high. (2) The value of external resistance is equal to the total internal resistance of all the cells. | |
| 1132. |
For what basic purpose the cells are connected (i) in series (ii) in parallel and (iii) in mixed grouping? |
| Answer» The cells are connected (i) in series, to get maximum voltage, (ii) in parallel, to get maximum current and (iii) in mixed grouping, to get maximum power. | |
| 1133. |
Do you know the blological connection of resistance and current to the human body? |
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Answer» When the skin is dry, resistance of humna body is `~~10^(5) ohm` , and when the skin is wet, the body resistance is `~~ 1500 ohm`. Our supply voltage is 220 V. Therefore, when the skin is dry, current through the body, `I=V/R = 220/10^(5) A= 2.2 mA` and current through the wet body, `I= V/R = 220/1500 A = 147.0 mA`. As is known, a current of 100 mA to 3 A through the human body is fatal , current of 20 mA to 100 mA causes paralysis , current of 10 mA to 20 mA causes muscular disorder and a current of 2 mA through the body causes tingling. Hence touching a 220 V live wire, especially when the body is wet, may have serious consequneces. |
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| 1134. |
How does the concept of change of resistance of a wire when strained (stretched) has been utilized in technology? |
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Answer» The concept of change of resistance of a wire when strained has been utilised by engineers (i) in providing a sense of touch to the robots. (ii) in making a machine called strain gauge sensor which is used to analyse the stresses and potential flaws in the structure of aircrafts and subnarines. This gauge is also used to measure the forces of blows in contact sports such as karate and football. (iii) In making a microscopic strain gauge fitted to a tiny catheter. This is used by cardiologists to measure blood pressure at precise locations within the heart. |
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| 1135. |
Why does Duracell company sell batteries in a package that includes a tester? |
| Answer» A tester in a duracell batteries package is a liquid crystal display the colour of which changes with temperature. It is used to test the strength of the battery. If the battery is fresh, the current supplied by it is strong, which will produce a large heating effect on the liquid crystal, resulting in a larger change in colour of the crystal. If the battery is weak, the change in colour of crystal in the tester is small. | |
| 1136. |
Internal resistance of a cell depends onA. distance between the electrodesB. concentration of electrolyteC. polarizationD. all of these |
| Answer» Correct Answer - D | |
| 1137. |
In an electrolytic solution the electric current is mainly due to the movement of free electrons. |
| Answer» An electrolyte solution is formed by mixing an electrolyte in a solvent. The electrolyte on dissolution furnishes ions. The preferred movement of ions under the influence of electric field is resoponsible for electric current. | |
| 1138. |
The value of internal resistance of ideal cell isA. zeroB. `0.5 Omega`C. `1 Omega`D. infinity |
| Answer» Correct Answer - A | |
| 1139. |
The resistance offered by electrolytic solution of a cell isA. impedanceB. reactanceC. internal resistance of cellD. admittance |
| Answer» Correct Answer - C | |
| 1140. |
As the current drawn from the cell increases, terminal potential difference of a cell isA. constantB. decreasesC. increasesD. none of these |
| Answer» Correct Answer - B | |
| 1141. |
The amount of work done by the cell in sending a unit charge once through the external resistance of the circuit isA. currentB. resistanceC. e.m.f. of a cellD. terminal potential difference of a cell |
| Answer» Correct Answer - D | |
| 1142. |
If 8 cells having an emf of 1.5 V are connected in series across a load having resistance of `10 Omega`. What is the current drawn by the load ? Assume internal resistance of all be `0.5 Omega`.A. 0.234 AB. 0.632 AC. 0.857 AD. None of these |
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Answer» Correct Answer - C Given, emf of each cell, E = 1.5 V Internal resistance of each cell, `r = 0.5 Omega` Number of cells, n = 8 Load resistance, `R = 10 Omega` Since, cells are connected in series `therefore " "` Total emf, `E_(T)=nE` `= 8xx1.5=12 V` Total resistance, `R_(T)=nr+R` `= (8xx0.5)+10` `= 14 Omega` Total circuit current, `I=(E_(T))/(R_(T))` `= (12)/(14)=0.857 A` |
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| 1143. |
On sending the current in the bridge, the bridge is said to be balanced, ifA. there is no defleaction in the galvanometerB. there is a deflection in the galvanometerC. partially deflection in the gavanometerD. none of these |
| Answer» Correct Answer - A | |
| 1144. |
(i) In a meter bridge , the balance point is found to be at `30 cm` from the end A when resistance `R` in left gap of bridge is of `12 Omega`. Find resistance `S` in the right gap of bridge (ii) If the cell and the galvanometer are interchanged in the balance point, would it effect flow of current through the galvanometer (iii) Calculate the balance point of the bridge if `R` and `S` are interchanged |
| Answer» There will be no deflection in the galvanometer as the condition of balanced bridge will still hold good. | |
| 1145. |
Wheaststone bridge is most sensitive when the resistance of all four arms areA. different orderB. same orderC. partially same orderD. partially different order |
| Answer» Correct Answer - B | |
| 1146. |
In the circuit given, the correct relation to a balanced Wheatstone bridge is A. `(P)/(Q) = (R)/(S)`B. `(P)/(Q) = (S)/(R)`C. `(P)/(R) = (S)/(Q)`D. None of these |
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Answer» Correct Answer - C |
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| 1147. |
In a Wheatstone bridge resistance connected the bridge is balanced , when the resistance are in the ratio |
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Answer» Wheatstone bridge is said to be balanced, when no current flows through the galavanometer arm of Wheatstone bridge. In balanced bridge, `P/Q = R/S` |
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| 1148. |
In the Wheatstone bridge network ABCD is balanced when `P = 500 Omega, Q = 250 Omega` and `S = 12 Omega`, what is the value of R ? A. `2 Omega`B. `4 Omega`C. `8 Omega`D. `6 Omega` |
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Answer» Correct Answer - D Given, `P = 500 Omega, Q = 250 Omega` and `S = 12 Omega` R = unknown resistance By balanced condition of Wheatstone bridge, we know `(P)/(Q)=(S)/(R )` Putting the values, `R = (Q xx S)/(P)=(250xx12)/(500)=6 Omega` `therefore` Value of unknown resistance, `R = 6 Omega` |
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| 1149. |
In a Wheatstone bridge resistance connected the bridge is balanced , when the resistance are in the ratio A. `(P)/(Q) = (R )/(S)`B. `(Q)/(P) = (S )/(R)`C. `PQ = RS`D. `(P)/(Q) = (S )/(R)` |
| Answer» Correct Answer - d | |
| 1150. |
Find the equivalent emf and internal resistance of the arrangement shown in figure. |
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Answer» Correct Answer - B `E_(eq)=(Sigma(E//r))/(Sigma(1//r))` `=((10//1)+(4//2)+(6//2))/((1//1)+(1//2)+(1//2))` `7.5V` `1/r=1/1+1/2+1/2` `:. R=0.5Omega` |
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