Ignoring power loss, power in primary = power in secondary. Power (P) = VI Primary voltage ($V_p$) = 220 V Secondary voltage ($V_s$) = 11000 V Power (P) = 88 W Secondary current ($I_s$) = ?
88 = 11000
$\implies i = \frac{88}{11 \times 10^3} = 8 \times 10^{-3} \text{ A}$
$\implies i = 8 \text{ mA}$
AB is a part of an electrical circuit (see figure). The potential difference \(V_A - V_B\), at the instant when current \(i = 2\) A and is increasing at a rate of 1 amp/second is: