Find resistor currents using KVL.
Resistive Circuit


R_1 and V_1 are parallel. So the voltage across R_1 is equal to V_1. This can be also calculated using KVL in the left hand side loop:

KVL for the left loop

-V_1+V_{R_1}=0 \rightarrow V_{R_1}=V_1=10V.
Now, use Ohm’s law to find I_{R_1}:

V_{R_1}=R_1 \times I_{R_1} \rightarrow I_{R_1}=\frac{V_{R_1}}{R_1}=0.1 A.
To find I_{R_2}, write KVL around the outer loop:
KVL for the outer loop
-V_1+V_{R_2}-V_2=0 \rightarrow V_{R_2}=V_1+V_2=20V.
Again, use Ohm’s law to determine I_{R_2}:

V_{R_2}=R_2 \times I_{R_2} \rightarrow I_{R_2}=\frac{V_{R_2}}{R_2}=0.2 A.

Now, tell me what is the current passing through V_1?

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        1. circuit
          10 \Omega and 3 \Omega are both short circuited. If you check the circuit you can easily see that both terminals are connected to each other for both.
          So, you have a circuit with a 10 V source and 2 \Omega and Ohm’s law says:
          I=\frac{V}{R}=5 A
          Let me know if you need more help.

          1. Thank you very much, sir.

            Surprisingly, in every guidebook I found the answer 7.5 Amp where there was no explanation behind the answer. But now I know the explanation and also know that correct answer is 5 Amp.

            I will post here if I need further help. Thanks again for your help.

        2. eliminate 10ohm resister becuase current will flow through low resistance patch then it is a series resistance circuit easy you can calculate otherwise let me know

    1. good question.
      Actually, it will be quite interesting. The voltage across R_1 is 10 V and therefore, the voltage drop on R_2 will be zero. You can also verify this by KVL around the outer loop:
      -10 V +V_{R_2}+10 V=0 \rightarrow V_{R_2}=0
      And using the Ohm’s law, we conclude that no current passes through R_2. The current of R_1 is the same as before because its voltage is not changed.

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