Question

   A 12 Ω and an 18 Ω are connected in parallel and the combination connected across 24oW dc line. (a) What is the resistance of the parallel combination? (b) What is the total current drawn by the combination?(c) what current is drawn by each resistor?

Answer 1

(a) Let's find the equivalent resistance of the parallel combination of the resistances:

$$\dfrac{1}{R_{eq}}=\dfrac{1}{R_{1}}+\dfrac{1}{R_2},$$ $$R_{eq}=\dfrac{R_1R_2}{R_1+R_2},$$ $$R_{eq}=\dfrac{12\ \Omega\cdot18\ \Omega}{12\ \Omega+18\ \Omega}=7.2\ \Omega.$$

(b) We can find the total current drawn by the combination from the Ohm's Law:

$$I_{tot}=\dfrac{V}{R_{eq}}=\dfrac{240\ V}{7.2\ \Omega}=33.3\ A.$$

(c) The voltage drop is the same across the each element in the parallel circuit ($V_1=V_2=V$). Therefore, we can find the current is drawn by each resistor:

$$I_1=\dfrac{V}{R_1}=\dfrac{240\ V}{12\ \Omega}=20\ A,$$ $$I_2=\dfrac{V}{R_2}=\dfrac{240\ V}{18\ \Omega}=13.3\ A.$$