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Circuit Theory

KCL and KVL

If the 12 Ω r...

Question

If the $12 Ω$ resistor draws a current of 1 A as shown in the figure, the value of resistance R is

Ω

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Solution

The correct option is B 6 $Ω$
Assuming voltage of the node $V_{a}$

$V_{a} = 1 \times 12 = 12 V$

Applying KCL,

$\Rightarrow - 2 + 1 + I = 0$
$I = 1$ A
$I = \frac{V_{a} - 6}{R}$
$= \frac{12 - 6}{R} = \frac{6}{R}$
$\Rightarrow I = \frac{6}{R}$
$\Rightarrow R = 6 Ω$

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If the 12 Ω resistor draws a current of 1 A as shown in the figure, the value of resistance R is

The correct option is B 6 Ω Assuming voltage of the node Va Va=1×12=12V Applying KCL, ⇒−2+1+I=0 I=1 A I=Va−6R =12−6R=6R ⇒ I=6R ⇒ R=6Ω

If the $12 Ω$ resistor draws a current of 1 A as shown in the figure, the value of resistance R is

The correct option is B 6 $Ω$
Assuming voltage of the node $V_{a}$

$V_{a} = 1 \times 12 = 12 V$

Applying KCL,

$\Rightarrow - 2 + 1 + I = 0$
$I = 1$ A
$I = \frac{V_{a} - 6}{R}$
$= \frac{12 - 6}{R} = \frac{6}{R}$
$\Rightarrow I = \frac{6}{R}$
$\Rightarrow R = 6 Ω$