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In the circuit shown, the potential difference between A and B is:
(1) 1 V
(2) 2 V
(3) 3 V
(4) 6 V
Given $E_1 = 1$ V, $E_2 = 2$ V, $E_3 = 3$ V, $r_1 = r_2 = r_3 = 1,\Omega$. $V_{AB} = \frac{E_1/r_1 + E_2/r_2 + E_3/r_3}{1/r_1 + 1/r_2 + 1/r_3} = \frac{1+2+3}{3} = \frac{6}{3} = 2$ V.
BETA
