Three resistors of $4\Omega$, $6\Omega$ and $10\Omega$ are connected in series across a $24\text{V}$ supply. Find (i) the equivalent resistance, (ii) the current in the circuit, (iii) the voltage drop across each resistor, and (iv) the total power dissipated in the circuit.

Step-by-step solution

1. (i) Add the resistances in series: $R_{\text{eq}}=4+6+10=20\Omega$.
2. (ii) Use Ohm’s law with the total voltage: $I=V/R_{\text{eq}}=24/20=1.2\text{A}$.
3. (iii) Find each voltage drop using $V=IR$.
4. - For $4\Omega$: $V_1=1.2\times 4=4.8\text{V}$.
5. - For $6\Omega$: $V_2=1.2\times 6=7.2\text{V}$.
6. - For $10\Omega$: $V_3=1.2\times 10=12.0\text{V}$.
7. (iv) Compute total power using $P=VI$.
8. $P_{\text{total}}=24\times 1.2=28.8\text{W}$.

Concept tested: Series circuit analysis

Published 4 October 2026 · How this solution was prepared and checked