Answer to Question #92942 in Electric Circuits for Efache

Question #92942
A battery of four cells each of e.m.f 1.5V and internal resistance 1.0ohms is connected to a 1ohm resistor in series with a parallel combination of two 2ohms resistors. Draw the circuit diagram and calculate:
a) the combined external resistance.
b) the current in the circuit
c) the lost volts in the battery
d) the current in one of the two resistors.
1
Expert's answer
2019-08-23T09:35:10-0400

The circuit diagram has the following outline:



Here we have: R1=1ΩR_1=1 \, \Omega , R2=2ΩR_2=2 \, \Omega , E=1.5V\Epsilon = 1.5 \,V, r=1Ωr = 1 \, \Omega


1) the combined external resistance is

R=R1+R22=1+1=2ΩR = R_1+\frac{R_2}{2} = 1 + 1 = 2 \, \Omega

2) the current in the circuit is

I=4E4r+R=41.54+2=1AI = \frac{4 \Epsilon}{4r + R}=\frac{4 \cdot 1.5}{4+2}=1 \,A

3) the lost volts in the battery are

U=4Ir=411=4VU=4 Ir=4 \cdot 1\cdot 1 = 4 \, V

4) the current through R2R_2 is

I2=I2=0.5AI_2 = \frac{I}{2} = 0.5 \,A


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Comments

Lauratu Aminu
05.03.24, 23:18

wow nice solution

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