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In the circuit shown , R_1=4Omega, R_2=R_3 =15Omega, R_4 = 30 Omegaand epsi = 10V .Calculate the equivalent resistance of the circuit and the current in each resistor. |
Answer» Solution : The GIVEN circuit can be redrawn as shown in Fig. Here `R_2 , R_3` and `R_4`are in parallel and their combined resistance R. will be `(1)/(R.) = (1)/(R_2)+ (1)/(R_3) + (1)/(R_4) = 1/15 + 1/15 + 1/30 = (2 + 2 + 1)/(30) = 5/30` ` rArr R. = 6OMEGA` Now R. and `R_1`are in series, hence equivalent resistance of the circuit `R = R. + R_1 = 6Omega +4Omega = 10 Omega` Main circuit CURRENT `I_1 = E/R = (10V)/(10 Omega) = 1 A ` `I_2+ I_3 + I_4 = I_1 = 1A` and potential difference between A and B remains constant, i.e., `15I_2 = 15I_3 = 30 I_4` ` rArr I_2 = I_3 = 2I_4` Therefore, on simplifying, we get `I_2 = I_3 = 0.4A` and `I_4 = 0.2 A ` |
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