Question

For the circuit shown next, calculate

(a) the power dissipated between points 1 and 2.

(b) the current drawn from the source.

(c) the voltage drop across resistor R_A.

(d) the voltage drop across resistor R_D.

(e) the potential difference between points 5 and 4.

Short answer

The power dissipated between points 1 and 2 is 47.25 mW.

The current drawn is 5.48 mA.

The voltage drop across R_A, R_D, and across points 5 and 4 is 11V, 4V, and 13V respaectively.

Step-by-step solution

Given Information

The following diagram with data is given;

Part (a)

The equivalent resistance between point's 1 and 2 is

$$\begin{gathered} R=\frac{R_B{R}_C}{R_B+R_C} \\ R=\frac{3\times 4}{3+4} \\ R=\frac{12}{7}\Omega \end{gathered}$$

The equivalent resistance across points 2 and 3 is:

$$\begin{gathered} V_{12}=\frac{R}{R_A+R+R\text{'}}V \\ {V}_{12}=\frac{12/7}{2+12/7+2/3}V \\ {V}_{12}=9V \end{gathered}$$

So the power dissipated between points 1 and 2 is:

P₁₂= V₁₂²/R

P₁₂= 9²/(12/7)

P₁₂= 47.25mW

Part (b)

The current drawn from the source is:

$$\begin{gathered} I=\frac{V}{R_A+R+R\text{'}} \\ I=\frac{24}{2+12/7+2/3}mA \\ I=5.48mA \end{gathered}$$

Part (c)

Voltage drop across R_A will be:

$$\begin{gathered} V_A=I{R}_A \\ {V}_A=5.48mA\times 2k\Omega \\ {V}_A=10.96V\approx 11V \end{gathered}$$

Part (d)

We know the voltage drop across R_A and voltage drop across points 1 and 2, so from Kirchoff's Voltage Law:

$$\begin{gathered} V_A+V_{12}+V_{23}=V \\ 11+9+V_{23}=24 \\ {V}_{23}=4V \end{gathered}$$

Thus the voltage drop across the points 2 and 3 is 4V.

Since the resistors R_D and R_E are connected in parallel across points 2 and 3, they will have the same voltage drop, equal to the voltage drop between 12 and 3 that is V₂₃.

Therefore, the voltage drop across resistor R_Dis 4V.

Part (e)

The potential difference across points 5 and 4 will be equal to the potential difference across points 1 and 3 since there is no electrical component between 1 and 5, and 3 and 4.

So,

$$\begin{gathered} V_{54}=V_{12}+V_{23} \\ {V}_{54}=9V+4V \\ {V}_{54}=13V \end{gathered}$$