Question

Four identical resistors of 4 ohm each are joined in circuit as shown in the figure. The cell B has emf 2 volts and its internal resistance is negligible. The ammeter will read :

• A. $\frac{3}{8}Ampere$
• B. $2Ampere$
• C. $\frac{1}{2}Ampere$
• D. $\frac{1}{8}Ampere$

Answer 1

The correct option is A $\frac{3}{8}Ampere$

Answer is A.

The given circuit diagram is redrawn to represent 3 resistors to be connected in parallel and the remaining 1 resistor to be connected in series with the parallel connection and the battery at the end.
Components connected in parallel are connected so the same voltage is applied to each component. In a parallel circuit, the voltage across each of the components is the same, and the total current is the sum of the currents through each component.
To find the total resistance of all components, the reciprocals of the resistances of each component is added and the reciprocal of the sum is taken. Hence, the Total resistance will always be less than the value of the smallest resistance.

That is, $\frac{1}{R_{total}}=\frac{1}{R_1}+\frac{1}{R_2}i.e.,R={R_{total}}^{-1}$.

In this case, the 3 resistance are connected in parallel and all three resistances are then connected with a 4 ohm resistance in series.

Components connected in series are connected along a single path, so the same current flows through all of the components. The current through each of the components is the same, and the voltage across the circuit is the sum of the voltages across each component. In a series circuit, every device must function for the circuit to be complete. One bulb burning out in a series circuit breaks the circuit. A circuit composed solely of components connected in series is known as a series circuit.The total resistance of resistors in series is equal to the sum of their individual resistances.

That is, $R_{total}=R_1+R_2$.
Now the total resistance is given as $\frac{4}{3}\Omega +4\Omega =\frac{16}{3}\Omega$.
The relation between resistance, current and voltage is given as $I=V/R$.
Therefore, the current through the circuit is $I=\frac{V}{R}=2\times \frac{3}{16}=\frac{3}{8}A$.