In the circuit shown in figure- R1-R2-R3-10 - Find the currents through R1 and R2-

As R_2=R_3 and V_A=V_B Potential difference across R_1 is zero. Hence, current through R_1⇒ i_1=0 and current through R_2 ⇒ i_2=V_ACR_2=10 V10 Ω=1 A ...

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Byju's Answer

Standard XIII

Physics

Kirchhoff's Voltage Law

In the circui...

Question

In the circuit shown in figure, $R_{1} = R_{2} = R_{3} = 10 Ω$ . Find the currents through $R_{1}$ and $R_{2}$ .

0 A, 1 A

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1 A, 0 A

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1.5 A, 1.5 A

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5 A, 5 A

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Solution

The correct option is A $0 A, 1 A$
As $R_{2} = R_{3}$ and $V_{A} = V_{B}$

Potential difference across $R_{1}$ is zero.
Hence, current through $R_{1} \Rightarrow i_{1} = 0$
and current through $R_{2}$

$\Rightarrow i_{2} = \frac{V_{A C}}{R_{2}} = \frac{10 V}{10 Ω} = 1 A$

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In the circuit shown in figure, R1=R2=R3=10 Ω. Find the currents through R1 and R2.

The correct option is A 0 A,1 AAs R2=R3 and VA=VB Potential difference across R1 is zero. Hence, current through R1⇒i1=0 and current through R2 ⇒i2=VACR2=10 V10 Ω=1 A

In the circuit shown in figure, $R_{1} = R_{2} = R_{3} = 10 Ω$ . Find the currents through $R_{1}$ and $R_{2}$ .

The correct option is A $0 A, 1 A$
As $R_{2} = R_{3}$ and $V_{A} = V_{B}$

Potential difference across $R_{1}$ is zero.
Hence, current through $R_{1} \Rightarrow i_{1} = 0$
and current through $R_{2}$

$\Rightarrow i_{2} = \frac{V_{A C}}{R_{2}} = \frac{10 V}{10 Ω} = 1 A$