Question

Two resistors, A and B, are connected in parallel across of a 6V battery. The current through B is found to be 2.0 A. When the two resistors are connected in series to the 6V battery, a voltmeter connected across the resistor A measures a voltage of 4V. Find the resistances of A and B.

Answer 1

Answer:

Resistance of A is $6\ \Omega$ and B is $3\ \Omega$

Explanation:

The voltage across both the resistances will be the same as they are connected in parallel.

V = Voltage = 6 V

$I_B=2\ \text{A}$

Resistance is given by

$R_B=\dfrac{V}{I_B}\\\Rightarrow R_B=\dfrac{6}{2}\\\Rightarrow R_B=3\ \Omega$

$V_B=V_b-V_A\\\Rightarrow V_B=6-4\\\Rightarrow V_B=2\ \text{V}$

Series connection

$V_A=4\ \text{V}$

The current is constant in series connection

$I=\dfrac{V_B}{R_B}\\\Rightarrow I=\dfrac{2}{3}\ \text{A}$

$R_A=\dfrac{V_A}{I}\\\Rightarrow R_A=\dfrac{4}{\dfrac{2}{3}}\\\Rightarrow R_A=6\ \Omega$

The resistance of A is $6\ \Omega$ and B is $3\ \Omega$.