InterviewSolution
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InterviewSolution
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Find out the folowing in the electric circuit given in (Fig. (a) Effective resistance of two `8 Omega` resistors in the combination (b) Current flowing through `4 Omega` resistor ( c) Potential difference across `4 Omega` resistance (d) Power dissipated in `4 Omega` resistor (e) Difference in ammeter readings, if any. . |
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Answer» (a) Since two `8 Omega` resistors are in parallel, their effective resistance `(R_p)` is given by `(1)/(R_p)=(1)/(8)+(1)/(8)=(1)/(4)` or `R_p = 4 Omega` (b) Total resistance in the circuit, `R = 4 Omega + R_p = 4 Omega + 4 Omega = 8 Omega` Current through the electric circuit, `I = (V)/( R) = (8 V)/(8 Omega) = 1 A` Since `4 Omega` resistor and `R_p` are in series, current through `4 Omega` resistors `= 1 A` (c) pd across `4 Omega` resistors, `V = IR = 1 xx 4 = 4 V` (d) Power dissipated in `4 Omega` resistors, `P = I^2 R = (1 A)^2 (4 Omega) = 4 W` (e) There is no difference in the readings of ammeters `A_1` and `A_2` as same current flows through all elements in a series curcuit. |
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