# Need help with physics problem?

A 75.0 ohm and 45 ohm resistor are connected in parallel and across a battery. The current is 0.294A. When the 45.0 ohm resistor is disconnected, the current from the battery drops to 0.116A. Determine the emf and the internal resistance of the battery.

I had thought you use the equation: E= V+ Ir but we have to determine r and the voltage we don't have. Im confused and if you can explain i'd appreciate it

### 6 Answers

- Andrew SmithLv 710 months agoFavorite Answer
This will be a simultaneous equation. You have two values so can solve for two unknowns.

You must set up equations with these unknowns.

OK what you are to solve for is V of the battery and R of the battery.

When the resistors are in parallel the effective resistance is 1/ ( 1/45 + 1/75)

V/ ( R + 1/(1/45 + 1/75) ) = 0.294

With only the single resistor you have

V/ (R+75) = 0.116

From this second one you can rearrange

V = 0.116(R + 75)

If you substitute this for V in the first equation it becomes

0.116(R+75) / (R + 1/(1/45 + 1/75) ) = 0.294

1/( 1/45+1/75) = 28.125

so substituting

0.116 (R+75) /(R + 28.125 ) = 0.294

Multiply both sides by ( R + 28.125)

0.116R + 75*0.116 = 0.294 * R + 0.294* 28.125

rearrange to get

75*0.116 - 0.294 * 28.125 = ( 0.294-0.116)R

divide both sides by ( 0.294-0.116)

R = ( 75* 0.116- 0.294* 28.125) / ( 0.294- 0.116) = 2.42 Ohms

Now substitute this back into V = 0.116( R +75)

= 0.116*(2.42 + 75) = 8.98 V

Check by putting both answers into the first equation.

8.98/(2.42+ 28.125) = 2.939925 ie it is very close to 2.94 The difference comes from the simplifying the results to three significant figures rather than keeping everything to limitless precision.

- VamanLv 710 months ago
I get V=13.78V and R internal resistance=45.865 Ohms. Check the correct values and pose the question.

- oldschoolLv 710 months ago
75||45 = 28.125Ω

We can write E = 0.294*(R+28.125)

and E = 0.116*(R+75)

Equate the equations for E: 0.294*(R+28.125) = 0.116*(R+75)

Re-arrange: R(0.294-0.116) = 75*0.116 - 28.125*0.294 = 0.43125

R = 0.43125/0.175 = 2.4228Ω

E = 0.294(2.4228+28.125) = 8.98V

Check

0.116(2.4228+75) = 8.98V checks

E ≈ 8.98V, <<<<<

R ≈ 2.42Ω <<<<<

- oubaasLv 710 months ago
Re = R1//R2 = 75*45/(75+45) = 3375/120 = 28,125 ohm

E /(r+Re) = 0.294

E = 0.294r+0.294*28.125 = 0.294r+8.269

E/(r+75) = 0.116

E = 0.116r+75*0.116 = 0.116r+8.700

E = 0.294r+8.269

E = 0.116r+8.700

0.294r+8.269 = 0.116r+8.700

0.178r = 0.431

r = 0.431/0.178 = 2.4228 ohm

E = 0.116*2.4228+8.7 = 8.9810 V

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- az_lenderLv 710 months ago
The equivalent resistance of the parallel combination is 1/(1/75 + 1/45) = 75*45/120

= 28.125 ohms, so the voltage drop across this resistor combination would be

(0.294A)(28.125 ohm) = 8.269 V.

When only the 45-ohm resistor is connected, the voltage drop across that resistor is 5.22 V.

Let R be the internal resistance of the battery.

It is necessary that

5.22V + 0.116A*R = 8.269V + 0.294A*R,

but this makes no sense, as it gives a negative number (around -11 ohms) for the internal resistance. No wonder you're confused, the problem itself describes a physically improbable situation.

- AlanLv 710 months ago
You have E and R in series for power supply.

Total Resistance = R_power_supply + (1/75 + 1/45)^(-1) = R_Power_supply + ((45 + 75)/ (45*75) )^(-1)

Total Resistance = R_power_supply + 28.125

E=IR = 0.294*(R_Power_Supply + 28.125) =0.294*R_Power_Supply +28.125*0.294

E = 0.294*R_Power_Supply + 8.26875

2nd situation

E = (0.116A) (R_Power_Supply + 75)

so 1st set two equations for E equal to find the R_Power_Supply

0.294*R_Power_Supply +8.26875 = 0.116*R_Power_Supply + 75*0.116

(0.294-0.116)R_Power_supply = 75*0.116 -8.26875 = 8.7 -8.26875 =

0.178*R_Power_Supply =0.43125

R_Power_Supply = 0.43125/ 0.178 = 2.422752809 Ohms

E or V power supply = 0.294( 2.422752809 +28.125) = 8.981039326 Volts

az_lender, you got it wrong. It was the 45 ohm resistor that was disconnected.