InterviewSolution
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InterviewSolution
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In the given circuit values are as follows epsi_1 = 2V, epsi_2 = 4V, R_1 = 1 Omegaand R_2 = R_3 = 1 Omega Calculate the currents through R_1 , R_2 and R_3 |
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Answer» Solution :Potential difference across BE is EQUAL to the e.m.f of `E_1`. i) The CURRENT through `R_1 ` is zero because the RESULTANT p.d across AG= 0 . The p.d. across BE due to `E_2 = -2 Omega` VOLTS and `E_1 = +2`volts, hence , p.d. across AG= +2 - 2 =0 ii) p.d. across `R_3 =-2` volts. `i_3 =(-2)/(R_3) = (-2)/1 = -2A`  iii) p.d. across BE = -2 volts. `i_2 = (-2)/(R_2) = (-2)/(1) = - 2A` Note :m You may solve this using Kirchoff.s Law as USUAL. |
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