Answer to Question #230493 in Electric Circuits for Angie

Question

Calculate (i) the equivalent resistance of the circuit shown below and (ii) the current flowing through each resistor. Resistances are R₁ = 100 Ω, R₂ = 50 Ω, R₃ = 50 Ω and R₄ = 75 Ω. The supply voltage is 6 V.        

Answer 1

Explanations & Calculations

• Dear Angie, you may have the corresponding figure of this question so you can follow my directions to the answer.

1)

• If you look carefully, you will see that the resistors $\small R_2,R_3 \,\&\, R_4$ are connected parallel to each other. (as the top end of the R4 resistor can be coincided with that of R2' s top end & th esame can be done with the bottom ed of R3 with R2)
• Then their equivalence is obtained by the method for parallel resistors,

$\qquad\qquad \begin{aligned} \small \frac{1}{R_{e1}}&=\small \frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}\\ \small R_{e1}&=\small 18.75\Omega \end{aligned}$

• Now you can draw a new circuit diagram with this equivalence & the remaining $\small 100\Omega(R_1)$ resistor & the battery.
• That arrangement is now series-connected hence the total equivalent resistance of the entire circuit is obtained easily as follows.

$\qquad\qquad \begin{aligned} \small R_{e-circuit}&=\small R_1+R_{e1}\\ &=\small \bold{118.75\,\Omega} \end{aligned}$

2)

• The full amount of current that flows in the circuit flows through R1 & only fractions of it are run through each other resistors depending on their resistor values.
• Now you know the total resistance or the eq. of the circuit. Thus you can simply calculate the total current of the circuit using $\small V=iR$ for the equivalent resistance.

$\qquad\qquad \begin{aligned} \small i_{total}=\small i_{R1}&=\small \frac{6V}{118.75\,\Omega}\\ &= \small \bold{0.0505\,A=50.5\,mA} \end{aligned}$

• Then, the voltage applied to the other 3 resistors is

$\qquad\qquad \begin{aligned} \small V_{r2,r3.r4}&=\small 6-i_{total}.R_1\\ &=\small 6-0.0505\times 100\Omega\\ &=\small 0.95V \end{aligned}$

• Then currents through each othe can be calculated,

$\qquad\qquad \begin{aligned} \small i_{R_2}&=\small i_{R3}=\frac{0.95V}{50\Omega}=0.019A=19mA\\\\ \small i_{R4}&=\small \frac{0.95}{75}=12.6mA \end{aligned}$