Two condensers of capacities 1 μ F and 2 μ F are connected in series and the system is charged to 120 volts. Then the potential difference on 1 μ F capacitor (in volts) will be

Question

Two condensers of capacities 1 μ F and 2 μ F are connected in series and the system is charged to 120 volts. Then the potential difference on 1 μ F capacitor (in volts) will be (A) 40 (B) 60 (C) 80 (D) 120

Tardigrade Admin · Accepted Answer

Charges developed are same, so C1 V1=C2 V2 ⇒ (V1/V2)=2 V1+V2=120 ⇒ V1=80 volts

Q. Two condensers of capacities $1 μ F$ and $2 μ F$ are connected in series and the system is charged to $120$ volts. Then the potential difference on $1 μ F$ capacitor (in volts) will be

40

60

80

120

Charges developed are same, so
$C_{1} V_{1} = C_{2} V_{2}$
$\Rightarrow \frac{V _{1}}{V _{2}} = 2$
$V_{1} + V_{2} = 120$
$\Rightarrow V_{1} = 80$ volts

Q. Two condensers of capacities 1μF and 2μF are connected in series and the system is charged to 120 volts. Then the potential difference on 1μF capacitor (in volts) will be

40

60

80

120

Charges developed are same, so C1​V1​=C2​V2​ ⇒V2​V1​​=2 V1​+V2​=120 ⇒V1​=80 volts

Q. Two condensers of capacities $1 μ F$ and $2 μ F$ are connected in series and the system is charged to $120$ volts. Then the potential difference on $1 μ F$ capacitor (in volts) will be

Charges developed are same, so
$C_{1} V_{1} = C_{2} V_{2}$
$\Rightarrow \frac{V _{1}}{V _{2}} = 2$
$V_{1} + V_{2} = 120$
$\Rightarrow V_{1} = 80$ volts