Question

6) You connect resistors of 1.0 , 2.0 12, and 3.0 22 across a battery as shown in the figure The current in the resistors is given by I., 12, and 13, respectively. If the current through the 3.0 12 resistor is 13 = 3.0 A, V una a) What is the current in the other two resistors? (3 points) b) What is the voltage across the three resistors? (2 points)

Answer 1

Let R1 = 1$\Omega$, R2 = 2$\Omega$, R3 = 3$\Omega$ and I3 = 3A. The circuit is redrawn below.

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(a) Applying Krichhoff's loop rule in the loop CDEFC, we get

$I_{2}R_{2}-I_{3}R_{3}=0$

$I_{2}R_{2}=I_{3}R_{3}$

$I_{2}=\frac{I_{3}R_{3}}{R_{2}}=\frac{3\times 3}{2}=4.5A$

So the current I2 is 4.5A.

Applying Krichhoff's loop rule in the loop GBCDEFG, we get

$I_{1}R_{1}-I_{3}R_{3}=0$

$I_{1}R_{1}=I_{3}R_{3}$

$I_{1}=\frac{I_{3}R_{3}}{R_{1}}=\frac{3\times 3}{1}=9A$

So the current I1 is 9A.

(b) Since the three resistors are in parallel, the voltage across one resistor is same as the voltage across all resistors. The voltage across R3 is

$V_{3}=I_{3}R_{3}=3\times 3=9V$

Therefore,

$V_{3}=V_{2}=V_{1}=V=9V$

So the voltage across the resistors is 9V.