a battery has an e.m.f of 6 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.when the switch is closed,the potential difference between the terminals of the battery is,involts?

Answer 1

explain how plzz

Answer 2

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.

Answer 3

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

Answer 4

0.8

Answer 5

How

Answer 6

Whats the answer

Answer 7

Better

Answer 8

Good