For the circuit both meters are idealized, the battery has no appreciable internal resistance, and the ammeter reads 1.25 . What does the voltmeter read? What is the emf of the battery? I tried this: 15 ohm and 10 ohm resistors are in series, hence equivalent to 25 ohm. Three parallel resistances 25 ohm, 15 ohm and 25 ohm have an equivalent resistance R is given by : 1/R = 1/25 + 1/15 + 1/25 = 11/75 or R = 75/11 ohm = 6.82 ohm 35 ohm, 45 ohm and 6.82 ohm are in series, hence equivalent to 35 + 45 + 6.82 = 86.82 ohm Current through the circuit = I = E/R = E/86.82 Amp (direction of current counter clockwise) Voltage drop across parallel combination of resistors = V = E - voltage drop across (45 + 35) ohm resistance. Hence, V = E - 80xE/86.82 = Ex6.82/86.82 = 0.078E Current through ammeter = Voltage across 25 ohm resistor/25 = 0.078E/25 = 1.25 (Given) Solving we get E = 400 Volt Voltage drop across 45 ohm resistor = 45xE/86.82 = 45x400/86.82 = 208 Volt but its wrong! Help Please!