Question

In the circuit shown, when $R$ is removed an additional resistance of $72\,$ ohm must be introduced in series with the battery in order to keep the current through $30\,$ ohm resistance unaltered. Hence $R$ is

Answer 1

Hint From the given circuit diagram, calculate the current, potential difference and the total current across each branch. Apply the formula below and substitute the calculated values to find the resistance of the resistor that is removed from the circuit.

Useful formula
The formula of the ohm’s law is given by
$V = IR$
Where $V$ is the potential difference, $I$ is the current flowing through the circuit and $R$ is the resistance.

Complete step by step solution
It is given that the
Resistor $R$ is removed from the circuit.
The additional resistor that must be connected to main the current, $72\,\Omega$

The value of the resistor, $R' = 30\,\Omega$
current $= \dfrac{2}{{28 + 30 + 72}} = \dfrac{1}{{75}}\,A$
Voltage across $CA =$ Total voltage - voltage across $AB$
${V_{CA}} = 2 - \dfrac{2}{5} = \dfrac{8}{5}\,V$
Total current $= \dfrac{8}{{5 \times 48}} = \dfrac{1}{{30}}\,A$
Since the resistor is connected in parallel, the potential difference developed across each resistor will be the same. Hence
$V = V'$
By substituting the ohms law in the above step,
$I'R = I\left( {30} \right)$
Rearranging the obtained equation, we get
$I' = \dfrac{{30I}}{R}$
Substituting the value of the resistance in the above equation, we get
$I' = \dfrac{{30}}{{75R}} = \dfrac{2}{{5R}}$
It is known that in a parallel circuit, the total current flows in the circuit will be equal to the sum of the currents flows in the individual branch.
$I = I + I'$
Substituting the values of the current in it,
$\dfrac{1}{{30}} = \dfrac{1}{{75}} + \dfrac{2}{{5R}}$
By simplify g the above equation by performing carious basic arithmetic operations,
$\dfrac{2}{{5R}} = \dfrac{1}{{50}}$
By further simplification of the above equation
$R = 20\,\Omega$

Hence the value of the resistance is obtained as $20\,\Omega$ .

Note
Remember that in a parallel circuit, the current flowing through the circuit will be the sum of the current flowing through each resistor and the voltage developed across each resistor will be the same. Hence the voltage across one resistor must be the subtraction of the other voltage with the total.