Answer to Question #153891 in Electric Circuits for Jerick

Question

Five resistors are connected to a 24 V resistor as shown.

(a) What is the total resistance of the circuit?

(b) How much current flows from the battery?

(c) How much power is supplied by the battery?

(d) What is the potential difference across each resistor?

(e) What is the current through each resistor?

(f) What is the power dissipated in each resistor?

Answer 1

Given that,

$V=24V\\ R_1=8\Omega\\ R_2=4\Omega\\ R_3=5\Omega\\ R_4=6\Omega\\ R_5=3\Omega\\ (a) R_1 \textsf{ and } R_2 \textsf{ are in parallel}. \textsf{Their combined resistance is given by,}\\ R_{12}=\frac{R_1×R_2}{R_1+R_2}=\frac{8×4}{8+4}=\frac{8}{3}.\\ R_{12} \textsf{ and } R_3 \textsf{ are in series}. \textsf{Their combined resistance is given by,}\\ R_{123}=R_{12}+R_3=\frac{8}{3}+5=\frac{23}{3}\\ R_{123} \textsf{ and } R_4 \textsf{ are in series}. \textsf{Their combined resistance is given by,}\\ R_{1234}=R_{123}+R_4=\frac{23}{3}+6=\frac{41}{3}\\ R_{1234} \textsf{ and } \textsf{ are in parallel }. \textsf{Their combined resistance is given by,}\\ R_{12345}= \frac{R_{1234}×R_5}{R_{1234}+R_5}=\frac{\frac{41}{3}×3}{\frac{41}{3}+3}=2.46\\ \textsf{ Hence, the total resistance of the circuit,} \\R_{Total}=2.46\Omega.\\ \hspace{2cm}\\$ $(b) \textsf{ From ohm's law,}\\ V=IR\\ I=\frac{V}{R}=\frac{24}{2.46}\\ I=9.76A\\ \hspace{2cm}\\$

$(c)P=IV=9.76×24=\\P=234.15W\\\\ \hspace{2cm}\\$

$(d) \textsf{ The p.d across } R_1\textsf{ and } R_2 \textsf{ are the same since they are connected in parallel}\\ \textsf{Likewise, the p.d across} R_4 \textsf{ and } R_5 are the same.\\ R_{12} \textsf{ is connected in series with} R_3 and R_{45}, hence, we divide the p.d across them.\\ R_{12}=\frac{8}{3}\Omega\\\\ \hspace{2cm}\\ R_{45}= \frac{R_4×R_5}{R_4+R_5}=\frac{6×3}{6+3}=2\Omega\\\\ \hspace{2cm}\\ R_3=5\Omega\\\\ \hspace{2cm}\\ \textsf{ p.d across }R_{12} =\frac{R_{12}}{R_{12}+R_{3}+R_{45}}×V\\\\ \hspace{2cm}\\ \textsf{ p.d across }R_{12}=\frac{\frac{8}{3}}{\frac{8}{3}+5+2}×24=6.62V\\\\ \hspace{2cm}\\ \textsf{ p.d across }R_{3}=\frac{R_3}{R_{12}+R_{3}+R_{45}}×V\\\\ \hspace{2cm}\\ \textsf{ p.d across }R_{3}=\frac{5}{\frac{8}{3}+5+2}×24=12.41V\\\\ \hspace{2cm}\\ \textsf{ p.d across }R_{45}=\frac{R_{45}}{R_{12}+R_{3}+R_{45}}×V\\\\ \hspace{2cm}\\ \textsf{ p.d across }R_{45}=\frac{2}{\frac{8}{3}+5+2}×24=4.97V.\\ \textsf{Hence, the p.d across the different resistors are given by,}\\ \textsf{ p.d across }R_1=6.62V\\ \textsf{ p.d across }R_2=6.62V\\ \textsf{ p.d across }R_3=12.41V\\ \textsf{ p.d across }R_4=4.97V\\ \textsf{ p.d across }R_5=4.97V\\ \hspace{2cm}\\$

$(e) \textsf{Since }R_{12}, R_3, R_{45}\textsf{ are in series, the same current 2.46A passes through them}\\ \textsf{The current 2.46A is then divided between} R_1\textsf{ and }R_2 as well as R_4 \textsf{ and } R_5 since they are connected in parallel.\\ I_{R_1}= \frac{R_2}{R_1+R_2}×I=\frac{4}{8+4}×2.46=0.82A\\\\ \hspace{2cm}\\ I_{R_2}=\frac{R_1}{R_1+R_2}×I=\frac{8}{8+4}×2.46=1.64A\\\\ \hspace{2cm}\\ I_{R_3}=2.46A\\\\ \hspace{2cm}\\ I_{R_4}=\frac{R_5}{R_4+R_5}×I=\frac{3}{6+3}×2.46=0.82A\\\\ \hspace{2cm}\\ I_{R_5}=\frac{R_4}{R_4+R_5}×I=\frac{6}{6+3}×2.46=1.64A\\ \hspace{2cm}\\$ $(f) P_{R_1}=I_{R_1}×\textsf{ p.d across }R_1\\=0.82×6.62=5.43W\\\\ \hspace{2cm}\\ P_{R_2}=I_{R_2}×\textsf{ p.d across }R_2\\=1.64×6.62=10.86W\\\\ \hspace{2cm}\\ P_{R_3}=I_{R_3}×\textsf{ p.d across }R_3\\=2.46×12.41=30.53W\\\\ \hspace{2cm}\\ P_{R_4}=I_{R_4}×\textsf{ p.d across }R_4\\=0.82×4.97=4.08W\\\\ \hspace{2cm}\\ P_{R_5}=I_{R_5}\textsf{ p.d across }R_5\\=1.64×4.97=8.15W$