Question 24.11: In Fig. 24.16, let R1 = 3 Ω, R2 = 6 Ω, R3 = 1 Ω, R4 = 7 Ω, a......
In Fig. 24.16, let R_{1} = 3 Ω, R_{2} = 6 Ω, R_{3} = 1 Ω, R_{4} = 7 Ω, and ΔV_{da} = V_{a} − V_{d} = 30 V. (a) What is the equivalent resistance between points a and d? (b) Evaluate the current passing through each resistor.
(a) We can simplify the circuit by the rule of adding resistances in series and in parallel in steps. The resistors R_{1} and R_{2} are in parallel and their equivalent resistance R_{12} between b and c is:
{\frac{1}{R_{12}}}={\frac{1}{R_{1}}}+{\frac{1}{R_{2}}}={\frac{1}{6\ \Omega}}+{\frac{1}{3\,\Omega}}={\frac{1}{2\,\Omega}}Then : R_{12}=2\Omega
Now R_{3}, R_{12}, and R_{4} are in series between points a and d. Hence, their equivalent resistance R_{eq} is:
R_{\mathrm{eq}}=R_{3}+R_{12}+R_{4}=1\,\Omega+2\,\Omega+7\,\Omega=10\,\Omega(b) The current I that passes through the equivalent resistor also passes through R_{3} and R_{4}. Thus, using Ohm’s law, we find that:
I=\frac{\Delta V_{d a}}{R_{\mathrm{eq}}}=\frac{30~V~}{10~\Omega}=3~\textrm{(Current through the battery,}R_{3}{\mathrm{~and~}}R_{4})Since \Delta V_{c b}=I R_{12}=I_{1}R_{1}=I_{2}R_{2}, then we find I_{1} and I_{2} as follows:
I_{1}={\frac{I R_{12}}{R_{1}}}={\frac{(3\,\mathrm{A})(2\,\Omega)}{3\,\Omega}}=2\,\mathrm{A}\ \ \mathrm{and}\ \ I_{2}={\frac{I R_{12}}{R_{2}}}={\frac{(3\,\mathrm{A})(2\,\Omega)}{6\,\Omega}}=1\,\mathrm{A}