# Homework Help: Circuit with ammeter , what does the voltmeter and battery read

1. Apr 29, 2008

### sonrie

For the circuit both meters are idealized, the battery has no appreciable internal resistance, and the ammeter reads 1.25 .

What is the emf of the battery?

I tried this:

15 ohm and 10 ohm resistors are in series, hence equivalent to 25 ohm.

Three parallel resistances 25 ohm, 15 ohm and 25 ohm have an equivalent resistance R is given by : 1/R = 1/25 + 1/15 + 1/25 = 11/75 or R = 75/11 ohm = 6.82 ohm

35 ohm, 45 ohm and 6.82 ohm are in series, hence equivalent to 35 + 45 + 6.82 = 86.82 ohm

Current through the circuit = I = E/R = E/86.82 Amp (direction of current counter clockwise)

Voltage drop across parallel combination of resistors = V = E - voltage drop across (45 + 35) ohm resistance. Hence,

V = E - 80xE/86.82 = Ex6.82/86.82 = 0.078E

Current through ammeter = Voltage across 25 ohm resistor/25 = 0.078E/25 = 1.25 (Given)

Solving we get E = 400 Volt

Voltage drop across 45 ohm resistor = 45xE/86.82 = 45x400/86.82 = 208 Volt but its wrong!

2. Apr 29, 2008

3. May 5, 2008

### sonrie

Can someone help me? need help!

4. May 5, 2008

### R A V E N

Are the correct values

$$V=206.1\;V$$

$$E=397.636\;V$$ ?

Last edited: May 5, 2008
5. May 5, 2008

### kamerling

I get V = 206.2 V and E = 397.9 V. probably just round off errors. Since the current is only given with 3 figures accuracy it should be V = 206 and E = 398, so we agree.
The origional poster should have used more digits in the intermediate calculations.

The problem is simper if you calculate the voltage drop across the 25 ohms resistance in series with the ammeter first.

6. May 5, 2008

### R A V E N

sonrie,you got the correct result more or less,althought you seem to complicated the procedure(I didn`t analyzed it throughoutly).

HINT:Calculate the voltage at the ends of branch with ammeter - the same voltage is at the end of other two branches.

Last edited: May 5, 2008