Find the current drawn from the battery and the potential difference between B and C.

0.2 A, 4 V

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1 A, 4 V

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2 A, 4 V

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0.2 A, 0.4 V

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Solution

The correct option is C
2 A, 4 V

Equivalent resistance, $R_{e q}$ = (2 || 2) + 1 + 1 + (4 || 4) = 1 + 1 + 1 + 2 = 5 $Ω$

So current in the circuit, I = $\frac{V}{R_{e q}}$ = $\frac{10 V}{5 Ω}$ = 2 A

Now voltage across BC, $V_{B C}$ = $I R_{B C}$ = $2 A \times (1 + 1) Ω$ = 4 V

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Similar questions

Q. The current flowing in the given circuit is

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. The potential difference between the points

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S. No	$[A]$	$[B]$	$[C]$	Rate of reaction
$1$	$0.2$	$0.2$	$0.2$	$2 \times 10^{- 4}$
$2$	$0.2$	$0.2$	$0.4$	$4 \times 10^{- 4}$
$3$	$0.2$	$0.4$	$0.2$	$2 \times 10^{- 4}$
$4$	$0.4$	$0.2$	$0.2$	$8 \times 10^{- 4}$

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Find the current drawn from the battery and the potential difference between B and C.

The correct option is C 2 A, 4 V Equivalent resistance, Req= (2 || 2) + 1 + 1 + (4 || 4) = 1 + 1 + 1 + 2 = 5 Ω So current in the circuit, I = VReq = 10 V5 Ω = 2 A Now voltage across BC, VBC = IRBC = 2 A ×(1+1) Ω = 4 V

The correct option is C
2 A, 4 V

Equivalent resistance, $R_{e q}$ = (2 || 2) + 1 + 1 + (4 || 4) = 1 + 1 + 1 + 2 = 5 $Ω$

So current in the circuit, I = $\frac{V}{R_{e q}}$ = $\frac{10 V}{5 Ω}$ = 2 A

Now voltage across BC, $V_{B C}$ = $I R_{B C}$ = $2 A \times (1 + 1) Ω$ = 4 V