Asked by komal yaseen

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?
Answered by komal yaseen
explain how plzz
Answered by Henry
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.
Answered by Anonymous
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
Answered by saqib
0.8
Answered by Anonymous
How
Answered by istifanus
Whats the answer
Answered by Tomy
Better
Answer
Good
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