Question

R1 = 6 ohm

R2 = 5 ohm

R3 = 4 ohm

R4 = 4 ohm

R5 = 8 ohm

with 10V

Solve for the voltage across resistor #5 (R5)?

Answer 1

Assume the circuit is as shown below

$R_2$ and $R_3$ are in series

$R_{2}+R_3=5+4=9\Omega$

$R_2,R_3and R_4$ combined gives

$1/R=1/9+1/4=13/36, R=36/13=2.769\Omega$

Effective resistance, $R_E=6+2.769+8=16.769\Omega$

given $v=10v$

$I=V/R_E=10/16.769=0.596A$

$V_{R5}=IR=0.596\times 8=4.768v$