## Question

###### 8 III A battery of 12v emf has internal resistance of 0.50 32 and is connected...

8 III A battery of 12v emf has internal resistance of 0.50 32 and is connected to a 6.00 load. What is the terminal Voltage of the battery under these Conditions and the in the external resistor ? + Hur {-128 power dissipeted r=o.sosh R= 6.002

## Answers

We have two reaistors in series.

The external resistor is

R = 6 ohms

Internal resistor is

r = 0.5 ohms.

So, total reaistance is

R+r = 6.5 ohms.

The voltage of the cell = 12 V

So, the current is given by

Current = voltage/Resistance.

I = 12/6.5 A (no need to divide here)

So, the voltage across the externla resistor, which will be equal to the terminal voltage is

Vt = IR = 12*6/6.5 = 11.077 V

The power dissipated in the external resistor is

P = I^2*R= 6*(12/6.5) ^2 = 20.45 Watts.

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