Question

A 40-Ω resistor is connected in series to the parallel combination of a 60-Ω resistor and a 30-Ω resistor to form a load a 120-V. Determine the current through and the voltage across each resistor.

Answer 1

To be given in question

$R_1=40\Omega\\R_2=60\Omega\\R_3=30\Omega$

V=120V

$R_{eq}=R_1+\frac{R_2R_2}{R_1+R_2}$

$R_{eq}=40+\frac{30\times60}{60+30}$

$R_{eq}=40+20=60\Omega$

$I_{net}=\frac{V}{R_{eq}}$

$I_{net}=\frac{120}{60}=2A$

$I_1=I_t=2A$

$I_2:I_3=\frac{1}{60}:\frac{1}{30}$

$I_2:I_3=1:2$

$I_2=\frac{1}{3}\times2=0.66A$

$I_3=\frac{2}{3}\times2=1.34A$

$V_1=I_1R_1=2\times40=80V$

$V_2=I_2R_2=0.66\times60=39.6V$

$V_3=I_3R_3=1.34\times30=40.2V$