In figure, charge on the capacitor is plotted against potential difference across the capacitor. The capacitance and energy stored in the capacitor are respectively.

Question

In figure, charge on the capacitor is plotted against potential difference across the capacitor. The capacitance and energy stored in the capacitor are respectively. (A) 12 μ F , 1200 μ J (B) 12 μ F , 600 μ J (C) 24 μ F , 600 μ J (D) 24 μ F , 1200 μ J

Tardigrade Admin · Accepted Answer

q = CV C =( q / V )=(120 × 10-6/10)=12 × 10-6 F C =12 μ F U =(1/2) CV 2 =(1/2) × 12 × 10-6 × 10 × 10 U =600 × 10-6 J U =600 μ J

Q. In figure, charge on the capacitor is plotted against potential difference across the capacitor. The capacitance and energy stored in the capacitor are respectively.

$12 μ F, 1200 μ J$

$12 μ F, 600 μ J$

$24 μ F, 600 μ J$

$24 μ F, 1200 μ J$

$q = C V$
$C = \frac{q}{V} = \frac{120 \times 1 0 ^{- 6}}{10} = 12 \times 1 0^{- 6} F$
$C = 12 μ F$
$U = \frac{1}{2} C V^{2}$
$= \frac{1}{2} \times 12 \times 1 0^{- 6} \times 10 \times 10$
$U = 600 \times 1 0^{- 6} J$
$U = 600 μ J$

Q. In figure, charge on the capacitor is plotted against potential difference across the capacitor. The capacitance and energy stored in the capacitor are respectively.

12μF,1200μJ

12μF,600μJ

24μF,600μJ

24μF,1200μJ

q=CV C=Vq​=10120×10−6​=12×10−6F C=12μF U=21​CV2 =21​×12×10−6×10×10 U=600×10−6J U=600μJ

$12 μ F, 1200 μ J$

$12 μ F, 600 μ J$

$24 μ F, 600 μ J$

$24 μ F, 1200 μ J$

$q = C V$
$C = \frac{q}{V} = \frac{120 \times 1 0 ^{- 6}}{10} = 12 \times 1 0^{- 6} F$
$C = 12 μ F$
$U = \frac{1}{2} C V^{2}$
$= \frac{1}{2} \times 12 \times 1 0^{- 6} \times 10 \times 10$
$U = 600 \times 1 0^{- 6} J$
$U = 600 μ J$