Consider the circuit shown below. Assume epsilon = 15 V, R_1 = 10 ohm, R_2 = 8 o
ID: 1414903 • Letter: C
Question
Consider the circuit shown below. Assume epsilon = 15 V, R_1 = 10 ohm, R_2 = 8 ohm, R_3 = 15 ohm, and R_4 = 18 ohm. What is the equivalent resistance of this circuit? How much current is drawn from the battery? Use Kirchoff's rules to find the current and voltage drop across each resistor. Suppose you wanted to minimize the current supplied by the battery, using the same four resistors. Sketch a new circuit diagram showing how you would configure the battery and resistors. Suppose you wanted to maximize the power output of your circuit, using the same four resistors. Sketch a new circuit diagram showing how you would configure the battery and resistors.Explanation / Answer
a)
R2 and R4 are in series.
Their equivalent resistance, R2+R4 = 8 + 18 = 26 Ohm
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R2+R4 is in parallel with R3
Their equivalent resistance is
R3||(R2+R4) = 15||26
= (15 x 26) / (15 + 26)
= 9.5 Ohm
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R3||(R2+R4) is in series with R1
The equivalent resistance is R1+[R3||(R2+R4)]
= 10 + 9.5
= 19.5 Ohm
Equivalent resistance of the circuit is 19.5 Ohm.
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b)
Current drawn = voltage of the battery / equivalent resistance
= 15 / 19.5
= 0.769 A
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c)
Consider that I current flows through the circuit. At B, it splits into I1 through R3 and I - I1 through R2 and R4.
Consider the loop with E, R3 and R1.
E - I1R3 - IR1 = 0
15 = 15I1 + 0.769 x 10
I1 = 0.487 A
Current through R2 and R4 is I - I1
= 0.769 - 0.487
= 0.282 A.
Voltage across R3 = I1R3
= 0.487 x 15 = 7.31 V
Voltage across R2 = (I - I1) R2
= 0.282 x 8 = 2.256 V
Voltage across R4 = (I - I1) R4
= 0.282 x 18 = 5.077 V.
Voltage across R1 = IR1
= 0.769 x 10 = 7.69 V
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d)
If we are connecting all the resistors in series with the battery, then resistance will be maximum and current will be minimum.