If 100 volts of potential difference is applied between a and b in the circuit of , find the potential difference between `c and d`.

Explanation

The charge distribution on different plates is shown in . Suppose charge `
`Q_1+Q_2`
` is given by the positive terminal of the battery, out of which `
`Q_1`
` resides on that positive plate of capacitor (1).and `
`Q_2`
` on that of (2) The remaining plates will have charges as shown in the figure. Take the potential at the point b to be zero. The potential at a will be 100 V. Let the potentials at points c and d be `
`V_c and `
`V_d`
` respectively. writing the equation Q = CV for the four capacitors, we get, `
`Q_1=6muF X100 V - 600mu`
`........(i) `
`Q_2 =6muF X (100 V - V _c)`
` ....... (ii) `
`Q_2 =6muF X(V_c - V_d)`
` .....(iii) `
`Q_2 = 6muF X V_d`
`......(iv) and from (ii) and (iii), `
`100 V- V_c = V_c -V_d`
` or, `
`2 v_c -V_d = 100 V`
` ....(v), and from (iii) and (iv), `
`V_c - V-d = V_d`
` or, `
`V_c = 2 V_d. ......(vi) From (v) and (vi), `
`V_d = (100)/(3) V and V_c = (200)/(3) V`
` that V_c - V_d = (100)/(3) V.