# 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?

## 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.

## To find the potential difference between the terminals of the battery when the switch is closed, we need to calculate the total resistance in the circuit.

The total resistance (R_total) can be found by summing the resistance of the resistor (R) and the internal resistance of the battery (r):
R_total = R + r

Given:
Internal resistance of the battery (r) = 0.4 ohm
Resistance of the resistor (R) = 2.6 ohm

Substituting the values:
R_total = 2.6 ohm + 0.4 ohm
R_total = 3 ohm

Now, we can use Ohm's Law to calculate the potential difference (V) across the terminals of the battery when the switch is closed. Ohm's Law states:
V = I * R_total

The current (I) flowing through the circuit can be determined using Ohm's Law again:
I = V / R_total

Given:
Electromotive force of the battery (emf) = 6 volts

Substituting the value of emf into the equation:
I = 6 volts / 3 ohm
I = 2 amperes

Now we can substitute the value of current (I) into the equation for potential difference (V):
V = (2 amperes) * (3 ohm)
V = 6 volts

Therefore, when the switch is closed, the potential difference between the terminals of the battery is 6 volts.