Question

The table below gives some known quantities for the pictured circuit.
What is the value of resistor $2$?

Battery Terminal Voltage	$12.0V$
Voltmeter $1$ Reading	$8.0V$
Ammeter $2$ Reading	$4.0V$

• A. $2.0ohm$
• B. $3.0ohm$
• C. $1.0ohm$
• D. $32.0ohm$
• E. $16.0ohm$

Answer 1

The correct option is C $1.0ohm$

Given : $E=12.0$ volts $V_1=8.0$ volts $I_2=4.0$ A

Thus potential difference across $R_2$ $V_2$ (or $V_3$) $=E-V_1=12.0-8.0=4.0$ volts

$\therefore$ Resistance of resistor 2, $R_2=\frac{V_2}{I_2}=\frac{4.0}{4.0}=1.0$ ohm