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How to find out the current through a circuit having resistances in series and parallel and having only on battery using kirchoff`s law

How to find out the current through a circuit having resistances in series and parallel and having only on battery using kirchoff`s law


2 Answers

Aarti Gupta
askIITians Faculty 300 Points
7 years ago

Series and parallel circuits are complex circuits.In other words, there will be parts with different voltages and parts with different currents. The rules for solving these circuits are the same , just apply the series rules for the series components and parallel rules for parallel components. These circuits can be very complex with many different voltage divisions and current divisions.
Kirchhoff's Current Law can be stated as "the sum of the currents entering a point must equal the sum of the currents leaving that point".Expression can be IT = I1 + I2 + . . . + IN

In the original circuit, on top , it can be seen that current flows from the negative end of the source (the bottom) counterclockwise around the circuit. It would flow up through R3 and then divide into I1 through R1 and I2 through R2, and then recombine into IT at the top before returning to the source. This makes R1 and R2 in parallel with each other. The total current (IT) and I3 would be the same and according to Kirchhoff's Current Law, I1 + I2 = IT. So this circuit has three different current values. Note since R1 and R2 are in parallel they would have the same voltage, but different currents.

To solve first find the equivalent resistance of R1 and R2 in parallel. Req = 1/(1/1200 + 1/2400) = 800 W. Now we can redraw the circuit and see that this R1-R2 equivalent is in series with R3. RT then is the sum of these two values. RT = 800 + 1200 = 2000 W. IT = V/RT = 12/2000 = 0.006 A or 6 mA.

The simplified circuit shows that there will be two voltage drops, one across the parallel combination and one across R3. These can be found as Veq = IReq = 0.006 x 800 = 4.8 V and V3 = IR3 = 0.006 x 1200 = 7.2 V. Kirchhoff's Voltage Law says that the two drops should add to equal VS and 4.8 V + 7.2 V = 12 V. So far so good.

Now that we know that V1 and V2 = 4.8 V, I1 and I2 can be found. I1 = V1/R1 = 4.8/1200 = 0.004 A or 4 mA, and I2 = V2/R2 = 4.8/2400 = 0.002 or 2 mA. Kirchhoff's Current Law says that these two currents should add to equal the current that they split from and 2 mA + 4 mA = 6 mA.

Thanks & Regards

Aarti Gupta

askiitians Faculty

Priya Bahuguna
8 Points
7 years ago
Thanku mam can u give me one more example. Of some complicated circuit

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