Physics

Physics

Problem

A series combination of n1n_1 capacitors, each of value C1C_1, is charged by a source of potential difference 4 V. When another parallel combination of n2n_2 capacitors, each of value C2C_2, is charged by a source of potential difference VV, it has the same (total) energy stored in it, as the first combination has. The value of C2C_2, in terms of C1C_1, is then: (2010 Pre) a. 16C1n1n2\frac{16C_1}{n_1 n_2} b. 2C1n1n2\frac{2C_1}{n_1 n_2} c. 16n2n1C116 \frac{n_2}{n_1} C_1 d. 2n2n1C12 \frac{n_2}{n_1} C_1
Intuitive explanation

The key is recognizing that capacitor energy depends on both capacitance and voltage through U=12CV2U = \frac{1}{2}CV^2. For the series combination, the total capacitance is Cseries=C1n1C_{\text{series}} = \frac{C_1}{n_1} charged to 4 V; for the parallel combination, it's Cparallel=n2C2C_{\text{parallel}} = n_2 C_2 charged to VV. Setting their energies equal gives you an equation where C2C_2 emerges as a ratio involving n1n_1, n2n_2, and the voltages—solve by equating 12Cseries(4)2=12CparallelV2\frac{1}{2}C_{\text{series}}(4)^2 = \frac{1}{2}C_{\text{parallel}}V^2.

Intuitive explanation

The key is recognizing that capacitor energy depends on both capacitance and voltage through U=12CV2U = \frac{1}{2}CV^2. For the series combination, the total capacitance is Cseries=C1n1C_{\text{series}} = \frac{C_1}{n_1} charged to 4 V; for the parallel combination, it's Cparallel=n2C2C_{\text{parallel}} = n_2 C_2 charged to VV. Setting their energies equal gives you an equation where C2C_2 emerges as a ratio involving n1n_1, n2n_2, and the voltages—solve by equating 12Cseries(4)2=12CparallelV2\frac{1}{2}C_{\text{series}}(4)^2 = \frac{1}{2}C_{\text{parallel}}V^2.

Browse the problem library