Two capacitors of capacitance 3 μ F and 6 μ F are charged to a potential of 12 V each. They are now connected to each other, with the positive plate of each joined to the negative plate of the other. The potential difference across each will be

Question

Two capacitors of capacitance 3 μ F and 6 μ F are charged to a potential of 12 V each. They are now connected to each other, with the positive plate of each joined to the negative plate of the other. The potential difference across each will be (A) 6 V (B) 4 V (C) 3 V (D) Zero

Tardigrade Admin · Accepted Answer

Here q1 = C1V = 3 × 10-6 ×12 = 36 μC and q2 = C2V= 6 × 10-6 × 12 = 72 μC Net charge = 72 - 36 = 36 μC This charge gets distributed among two capacitors. ∴ V1 = (12/3) = 4V and V2 = (24/6) = 4 V

Q. Two capacitors of capacitance $3 μ F$ and $6 μ F$ are charged to a potential of $12 V e a c h$ . They are now connected to each other, with the positive plate of each joined to the negative plate of the other. The potential difference across each will be

6 V

4 V

3 V

Zero

Here $q_{1} = C_{1} V = 3 \times 1 0^{- 6} \times 12$ = 36 $μ$ C and
$q_{2} = C_{2} V = 6 \times 1 0^{- 6} \times 12$ = 72 $μ$ C
Net charge = 72 - 36 = 36 $μ$ C
This charge gets distributed among two capacitors.
$∴$ $V_{1} = \frac{12}{3}$ = 4V and $V_{2} = \frac{24}{6}$ = 4 V

Q. Two capacitors of capacitance 3μF and 6μF are charged to a potential of 12Veach. They are now connected to each other, with the positive plate of each joined to the negative plate of the other. The potential difference across each will be

6 V

4 V

3 V

Zero

Here q1​=C1​V=3×10−6×12 = 36 μC and q2​=C2​V=6×10−6×12 = 72 μC Net charge = 72 - 36 = 36 μC This charge gets distributed among two capacitors. ∴ V1​=312​ = 4V and V2​=624​ = 4 V

Q. Two capacitors of capacitance $3 μ F$ and $6 μ F$ are charged to a potential of $12 V e a c h$ . They are now connected to each other, with the positive plate of each joined to the negative plate of the other. The potential difference across each will be

Here $q_{1} = C_{1} V = 3 \times 1 0^{- 6} \times 12$ = 36 $μ$ C and
$q_{2} = C_{2} V = 6 \times 1 0^{- 6} \times 12$ = 72 $μ$ C
Net charge = 72 - 36 = 36 $μ$ C
This charge gets distributed among two capacitors.
$∴$ $V_{1} = \frac{12}{3}$ = 4V and $V_{2} = \frac{24}{6}$ = 4 V