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# Electric Circuits watch

1. I've been stuck on these two questions for 90 minutes now and I'm no nearer to completing it. Can I have some help on how to do these please?

(a) A battey is used to drive two different load resistors. When it is connected to a 20 ohm resistor a current of 0.10 A is drawn, whereas a load resistor of 80 ohm resistor draws 0.033 A. Determine the e.m.f. and the internal resistance.

(b) You have a moving coil meter of full scale deflection 50mA and resistance 5 ohms. Explain how you would use it to measure voltages up to 10V determining the values of any components you have used.
3. Just write down what you know and it should become obvious how to proceed.

(a)
0.1 = E/(20+r)
0.033 = E(80+r)

You now have two equations with two unknowns, so can solve simultaneously.

(b)
Just write down what you know again.

0.05 = 10/(5+R)

Solve for R and you know that when placed in series with that resistor R, a 10V excitation will cause full scale deflection since 50mA flows.
4. (Original post by EdwardCurrent)
Just write down what you know and it should become obvious how to proceed.

(a)
0.1 = E/(20+r)
0.033 = E(80+r)

You now have two equations with two unknowns, so can solve simultaneously.

(b)
Just write down what you know again.

0.05 = 10/(5+R)

Solve for R and you know that when placed in series with that resistor R, a 10V excitation will cause full scale deflection since 50mA flows.
^^This^^

To OP just in case: if you didn't know already, the formula used here was

E=I(R-r)

'E' is e.m.f and 'r' is the internal resistance.

just rearrange to make I the subject, and Bob's your Uncle.
5. thanks for that, I've managed to solvethe question now

phew, frustrations gone

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