Answer to Question #179016 in Electric Circuits for Dylan

Question

a circuit with a 2 cell battery, and a resistor(R1) connected in series. Also 2 resistors in parallel at one point in the circuit(R2 & R3) and another 2 resistors in parallel at another point(R4 & R5). V1 = 4.0V, V2 = 1.5V , I2 = 0.2A , I1 = 0.7A, I4 = 0.3A, and R4 = 70Ω. Find all other voltages, currents, resistances and the totals for the circuit

Answer 1

Explanations & Calculations

• Refer to the attached figure.

1. Since the current through the series resistor R1 represents the total current, then the total current is $\small I=i_1 =0.7A$
2. Applying V=iR to R1,

$\qquad\qquad \begin{aligned} \small R_1&=\small \frac{V}{i_1}=\frac{4V}{0.7A}=\bold{5.7\Omega} \end{aligned}$

3.Applying $\small V=iR$ to R2,

$\qquad\qquad \begin{aligned} \small R_2&=\small \frac{V}{i_2}=\frac{1.5V}{0.2A}=\bold{7.5\Omega} \end{aligned}$

4.Current flow through R3 is

$\qquad\qquad \begin{aligned} \small i_3&=\small I-i_2=0.7-0.2=\bold{0.5A} \end{aligned}$

5.Since R2 & R3 are parallel to each other, they experience the same potential drop across them.

Therefore,

$\qquad\qquad \begin{aligned} \small V_3&=\small V_2=\bold{1.5V} \end{aligned}$

6.Then applying $\small V=iR$ to R3,

$\qquad\qquad \begin{aligned} \small R_3&=\small \frac{V}{i}=\frac{1.5V}{0.5A}=\bold{3\Omega} \end{aligned}$

7.Applying $\small V=iR$ to R4,

$\qquad\qquad \begin{aligned} \small V_4 &=\small iR=0.3A\times70\Omega=\bold{21V} \end{aligned}$

8.Since R5 is connected in parallel to R4 they both experience the same potential drop. Then,

$\qquad\qquad \begin{aligned} \small V_5&=\small \bold{21V} \end{aligned}$

9.The current through R5 is

$\qquad\qquad \begin{aligned} \small i_5&=\small 0.7-0.3=\bold{0.4A} \end{aligned}$

10.Applying $\small V=iR$ to R5,

$\qquad\qquad \begin{aligned} \small R_5&=\small \frac{V}{i}=\frac{21V}{0.4A}=\bold{52.5\Omega} \end{aligned}$

11.Total voltage provided by the battery hence the potential of the battery.

$\qquad\qquad \begin{aligned} \small V_b&=\small 4V+1.5V+21V=\bold{26.5V} \end{aligned}$

12.In case of an equivalent resistance that is external to the circuit,

$\qquad\qquad \begin{aligned} \small R_e&=\small R_1+\frac{R_2.R_3}{R_2+R_3}+\frac{R_4.R_5}{R_4+R_5}\\ &=\small 5.7+\frac{7.5\times3}{7.5+3}+\frac{70\times52.5}{70+52.5}\\ &=\small 5.7+2.14+30\\ &=\small \bold{37.84\Omega } \end{aligned}$

Which can also be calculated by

$\qquad\qquad \begin{aligned} \small R_e &=\small \frac{26.5V}{0.7A}=37.84\Omega \end{aligned}$