When two capacitors of capacities 3mu F and 6mu F are connected in series and connected to 120 V- the potential difference across 3mu F is-40V80V180V60V

Question

Toppr Admin · Accepted Answer

C-3-xD7-69-2-x3BC-FV-120VQ-CV-120-xD7-2-240-x3BC-CV3-Q3-x3BC-F-240-x3BC-C2-x3BC-F-80V

When two capacitors of capacities $3 μ F$ and $6 μ F$ are connected in series and connected to $120 V,$ the potential difference across $3 μ F$ is:
$40 V$
$80 V$
$180 V$
$60 V$

$C = \frac{3 \times 6}{9} = 2 μ F$
$V = 120 V$
$Q = C V = 120 \times 2 = 240 μ C$
$V_{3} = \frac{Q}{3 μ F} = \frac{240 μ C}{2 μ F} = 80 V$

When two capacitors of capacities 3μF and 6μF are connected in series and connected to 120V, the potential difference across 3μF is:40V80V180V60V

C=3×69=2μFV=120VQ=CV=120×2=240μCV3=Q3μF=240μC2μF=80V

When two capacitors of capacities $3 μ F$ and $6 μ F$ are connected in series and connected to $120 V,$ the potential difference across $3 μ F$ is:
$40 V$
$80 V$
$180 V$
$60 V$

$C = \frac{3 \times 6}{9} = 2 μ F$
$V = 120 V$
$Q = C V = 120 \times 2 = 240 μ C$
$V_{3} = \frac{Q}{3 μ F} = \frac{240 μ C}{2 μ F} = 80 V$