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Ohm's Law

In the circui...

Question

In the circuit shown in figure, $R_{1} = R_{2} = R_{3} = R$
Table 1 Table - 2
(a) current through $R_{1}$ (p) E/R
(b) current through $R_{2}$ (q) 2E/R
(c) current through $R_{3}$ (r) E/2R
(s) Zero

a - p ; b - p ; c - p

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a - p ; b - q ; c - s

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a - q ; b - r ; c - r

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a - r ; b - s ; c - p

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Solution

The correct option is A a - p ; b - p ; c - p
Voltage across $V_{R_{1}} = E$
Therefore $I_{R_{1}} = \frac{E}{R_{1}} = \frac{E}{R}$
Similarly, for $R_{2}$ :
Voltage across $V_{R_{2}} = E$
Therefore $I_{R_{2}} = \frac{E}{R_{2}} = \frac{E}{R}$
Voltage across $R_{3}$ $V_{R_{3}} = E + E - E = E$
Thus $I_{R_{3}} = \frac{V_{R_{3}}}{R_{3}} = \frac{E}{R}$
Therefore option (a) is correct.

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R_{1} = R_{2} = R_{3} = 10 Ω

R_{1}

R_{2}

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