Question

A battery has an e.m.f. of 6.0 volts and an internal resistance of 0.4 ohm.it is
connected to a 2.6 ohm resistor through a SPST (single pole, single throw) switch.
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Answer 1

Explanation:

E = 6 Volts

r = 0.4 Ohms

R = 2.6 Ohms.

I = E/(r+R) = 6/(0.4+2.6) = 2 Amps. =

Current flowing in the circuit.

Vt = I*R = 2 * 2.6 = 5.2 Volts. = Potential difference between terminals

of the battery.

Vr = I * r = 2 * 0.4 = 0.8 Volts. =

Voltage lost across the internal resistance.

Vt + Vr = 5.2 + 0.8 = 6.0 Volts = Battery e.m.f.

a battery has an emf of 6.0 volts and an internal resistance of 0.4 ohms. its is connected to a 2.6 ohms resistor a switch. when switch is open the potential difference between the terminals of the battery is 0.8