Question

A flashlight contains a battery of two cells in series, with a bulb of resistance 12 Ohms. The internal resistance of each cell is 0.26 ohms. If the potential difference across the bulb is 2.88V, what is the emf of EACH cell?

Answer 1

1.5024

Step-by-step explanation:

Draw a diagram. Put the two cells in series. Now draw 3 resistors. Two of them equal 0.26 ohms each. The third one is the lightbulb which is 12 ohms.

R = 0.26 + 0.26 + 12 = 12.52

The bulb has a voltage of 2.88 volts across it. You can get the current from that.

i = E / R

i = 2.88 / 12 =

i = 0.24 amps.

Now you can get the voltage drop across the two cells.

E = ?

R = 0.26

i = 0.24 amps

E = 0.26 * 0.24

E = 0. 0624

Finally divide the 2.88 by 2 to get 1.44

Each cell has an emf of 1.44 + 0.0624 = 1.5024