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

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

Step-by-step 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=(V)/(I_B)\\\Rightarrow R_B=(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=(V_B)/(R_B)\\\Rightarrow I=(2)/(3)\ \text{A}`

`R_A=(V_A)/(I)\\\Rightarrow R_A=(4)/((2)/(3))\\\Rightarrow R_A=6\ \Omega`

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