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(i) The circuit is shown in figure.

(ii) Suppose I₁ are I₂ current drawn from cells ε₁ and ε₂ respectively, then according to Kirchhoff’s junction law, current in R = 5Ω I = I₁ + I₂.

Applying Kirchhoff’s second law to mesh ABFEA

1 × I₁ + 1.5 – 5(I₁ + I₂) = 0

⇒ 6I₁ + 5I₂ = 1.5 …(i)

Applying Kirchhoff’s second law to mesh CDEFC

–2I₂ + 2 – 5(I₁ + I₂) = 0

⇒ 5I₁ + 7I₂ = 2 …(ii)

Solving equation (i) and (ii), we get

\(I_1=\frac{1}{34}A,I_2=\frac{9}{34}A\)

\(I=I_1+I_2=\frac{1}{34}+\frac{9}{34}=\frac{10}{34}A\)

Potential difference across R = 5Ω resistor

\((I_1+I_2)R=\frac{10}{34}\times5=\frac{25}{17}\) volt

The emf of a cell is equal to the terminal voltage, when the circuit is open. 

The emf of a cell is less than the terminal voltage, when the cell is being charged, i.e., V = E + ir

(i) In an open circuit, the emf of a cell and terminal voltage are same.

(ii) In closed circuit, a current is drawn from the source, so, V = E – Ir, it is true/valid, because each cell has some finite internal resistance.

Because the cell has some finite internal resistance. An emf is determined when the cell is in open circuit and no current is drawn .

No.

When electrons drift in a metal from lower to higher potential, does it mean that all the free electrons of the metal are moving in the same direction ?

(i) Water in the waterfall flows from a higher level to the lower level because of gravitational attraction.

(ii) In an electric circuit, electrons flow from a point of lower potential to the point of higher potential.

(iii) The difference between the electrostatic potential of the positive end and the negative end of an electric cell is the potential difference of the cell.

(iv) Three electric cells of potential difference 1.5 V each have been connected as a battery. The potential difference of the battery will be 4.5 V.

(v) An electric current flowing in a wire creates magnetism around the wire.

(i) 6V 

(ii) 8V. 

Note: In (i), three cells are connected in series. 

∴ Total potential difference = 2V + 2V + 2V = 6V. 

In (ii), four cells are connected in series. 

∴ Total potential difference = 2V + 2V + 2V + 2V 

= 8 V

Water flows from a certain height of a mountain towards the ground.

If a coil is wound around an iron screw (or an iron rod) and a current is passed through the coil with a cell (or a battery), the screw behaves as a magnet as long as there is current in the coil. The i system of the coil and the screw is called an electromagnet. It is an example of magnetic effect of electric current. Applications: Electric bell, cranes for moving heavy loads from one place to another, toys that run on electric cells.

Due to the elastic iron strip, the making and breaking of the circuit occur alternately and the bell continues to ring as long as the key in the circuit is closed.

The strength of an electromagnet can be increased by increasing the current producing the magnetic field.

Current is the measure of rate of flow of charge. Therefore the rate of flow of charge through R₂ is also I₁. 

Correct option is (C) 60 cm

(a) Total Resistance in series:

R =  R₁ + R₂ + R₃ 

(b) Total Resistance in parallel:

1/R =  1/R₁ + 1/R₂ + 1/R₃ 

Data : G = 500 Ω, 

V_g = 5 V, 

V = 100 V 

To increase the range of the voltmeter by a factor p = \(\cfrac{V}{V_g}\), a resistance R should be connected in series with it.

R_s = G(p – 1) = 500\(\left(\cfrac{100}5-1 \right)\)

= 9500 Ω

Data : G = 60 Ω, 

I_g = 50 mA = 5 × 10^-2 A, 

I = 5 A, 

V = 50 V

(i) S = \(\cfrac{GI_g}{I-I_g}\)

= \(\cfrac{60\times5\times10^{-2}}{5-5\times10^{-2}}\) = \(\cfrac3{4.95}\) = 0.6061 Ω

A resistance of 0.6061 Ω should be connected in parallel to the coil of the galvanometer to measure current up to 5 A.

(ii) R_S = \(\cfrac{V}{I_g}\) – G 

= \(\cfrac{50}{5\times10^{-2}}\) – 60 

= 1000 – 60 = 940 Ω

A resistance of 940 Ω should be connected in series with the coil of the galvanometer to measure voltage up to 50 V.

Data : G = 297 Ω, S = 3 Ω

I_g = \(\cfrac{S}{S+G}\). I

∴ \(\cfrac{I_g}I\) = \(\cfrac{S}{S+G}\) = \(\cfrac{3}{3+297}\) = \(\cfrac{3}{300}\) = 0.01

This is the fraction of the current through the galvanometer.

[Note: The fraction of the current through the shunt

= \(\cfrac{I_s}I\) = 1 - 0.01 = 0.99]

Positive electrode is known as Anode.

B. electrical, mechanical, light, heat, magnetic, and chemical

r=0 when short-circuited r 

=innity(high) when open circuited.

No, ohm’s law is not obeyed by all the elements. For example, vaccum diode tube and semiconductor diode.

Correct option is (A) \(\cfrac{S}{S+G}\)

\(I=\frac{E+E}{R+r_1+r_2}\)

V₁ = E - Ir₁ = E - \(\frac{2E}{r_1+r_2+R}r_1=0\)

or  \(E=\frac{2Er_1}{r_1+r_2+R}\)

\(\Rightarrow\) r₁ + r₂ + R = 2r₁

\(\Rightarrow\) R = r₁ - r₂

Current flowing in the potentiometer wire

\(I=\frac{E}{R_{total}}=\frac{2.0}{15+10}=\frac{2}{25}A\)

Potential difference across the wire \(\frac{2}{25}\times10=\frac{20}{25}=0.8V\)

Potential gradient \(k=\frac{V_{AB}}{l_{AB}}=\frac{0.8}{1.0}=0.8V/m\)

Now, current flowing in the circuit containing experimental cell,

\(\frac{1.5}{1.2+0.3}=1A\)

Potential difference across length AO = 0.3 x 1 = 0.3 V

Length AO = \(\frac{0.3}{0.8}m=\frac{0.3}{0.8}\times100 \,cm=37.5\,cm\) 

Potential gradient is defined as the potential difference (the fall of potential from the high potential end) per unit length of the wire.

Sliding the jockey on the potentiometer wire decreases the cross sectional area of the wire and thereby affects the fall of potential along the wire. This affects the potentiometer readings. Flence, the jockey should not be slided along the potentiometer wire.

Power \(P=\frac{V^2}{R}\,\infty\,\frac{1}{R}\) for same voltage, the bulbs being in parallel.

\(\frac{P_1}{P_2}=\frac{R_2}{R_1}=\frac{2}{1}.\)

Thus, ratio of power dissipated is 2 : 1.

(a) When both S₁ and S₂ are dosed, the effective resistance of the circuit decreases and hence, current will increase.

(b) When both S₁ and S₂ are open, the effective resistance of the circuit increases and hence, current will decrease.

(c) When S₁ is closed and S₂ is open, the effective resistance of the circuit decreases and hence current will increase. 

[Current will be more than case (b) but less than in case (a)]

(b) Yellow – Violet – Orange – Silver

Current is a scalar quantity because it does not obey the laws of vector addition.

Current density is a vector quantity.

(c) He should change S to 3Ω and repeat the experiment. 

Wheatstone bridge is an application of Kirchhoff’s Law.

Metre bridge is a form of Wheatstone’s bridge.

(b) The battery of potentiometer can have a voltage of 15V and R adjusted so that the potential drop across the wire slightly exceeds 10V. 

If galvanometer and battery are interchanged, there is no effect on the balance Point3.

If the galvanometer and battery are also interchanged there will be no effect on the position of the balance point

this is based on the principle of meter bridge or WHEATSTONE BRIDGE

according to tho this principle If there is null deflection in the galvanometer then the Wheatstone bridge is in balanced condition and in this condition

P/Q=R/S

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In Kelvin’s method of determination of the galvanometer resistance using a Wheatstone metre bridge, the galvanometer is connected in one gap of the bridge and a variable known resistance is connected in the other gap. The junction of the two gaps (say, B) is connected directly to a pencil jockey.

The jockey is tapped along the wire to locate the equipotential balance point D when the galvanometer shows no change in deflection.

Since the galvanometer shows the same deflection on making or breaking the contact between the jockey and the wire, the method is an equal deflection method.

Kelvin’s method of determination of the galvanometer resistance is an equal deflection method. The balance point in Kelvin’s method is a point on the wire for which the bridge network is balanced and the galvanometer shows no change in deflection.

For manganin, the temperature coefficient of resistance is very low and its resistivity is quite high. Due to it, the resistance of manganin wire remains almost unchanged with change in temperature. Hence it is used. 

1. Wheatstone’s metre bridge is used for measuring the values of very low resistance precisely. 

2. It can also be used to measure the quantities such as strain galvanometer, resistance, capacitance of a capacitor, inductance of an inductor, impedence of a combination of a resistor, capacitor and inductor and the internal resistance of a cell.

j = σ E Where σ = conductivity and E = electric field. 

The resistance of a material depends upon temperature, strain, humidity, etc.

[Note : Depending upon the factors stated above, resistance may vary from near zero to thousands of megaohm.]

Resistance of a conductor depends on (1) length of the conductor (2) Area of cross section of the conductor. 

The potentiometer has the advantage that it draws no current from the voltage source being measured. 

Potentiometer. 

Functions of the series, resistance: 

1. It increases the effective resistance of the voltmeter. 

2. It drops off a larger fraction of the measured potential difference thus protecting the sensitive meter movement of the basic galvanometer.

3. With resistance of proper value, a galvanometer can be modified to a voltmeter of desired range. .

Functions of the shunt:

1. It lowers the effective resistance of the ammeter 

2. It is used to divert to a large part of total current by providing an alternate path and thus it protects the instrument from damage. 

3. With a shunt of proper value, a galvanometer can be modified into an ammeter of practically any desired range.

If R is increased, the current through the wire will decrease and hence the potential gradient will also decrease, which will result in increase in balance length. So J will shift towards B.