Question

analyze the circuit to determine the unknown values.
use the calculated resistance on the source row

r (ω)

i (a)

v (v)

p (w)

----

----

----

----

----

a

b

c

d

source

circuit with a 9v power supply, resistor a: 30ω, resistor b: 30ω, resistor c: 40ω, resistor d: 20ω, where a, b, c are in parallel, and this parallel combination is in series with d

Explanation:

Step1: Find parallel resistance (a,b,c)

First, calculate the equivalent resistance of the parallel branch (resistors a, b, c):
$$\frac{1}{R_{parallel}} = \frac{1}{30\Omega} + \frac{1}{30\Omega} + \frac{1}{40\Omega} = \frac{4 + 4 + 3}{120\Omega} = \frac{11}{120\Omega}$$
$$R_{parallel} = \frac{120}{11} \approx 10.91\Omega$$

Step2: Find total circuit resistance

Add resistor d to the parallel resistance for total circuit resistance:
$$R_{total} = R_{parallel} + 20\Omega = \frac{120}{11}\Omega + 20\Omega = \frac{120 + 220}{11}\Omega = \frac{340}{11} \approx 30.91\Omega$$

Step3: Find total current (I_d)

Use Ohm's Law for total circuit current (this is the current through resistor d):
$$I_d = \frac{V_{total}}{R_{total}} = \frac{9V}{\frac{340}{11}\Omega} = \frac{99}{340} \approx 0.291A$$

Step4: Find voltage across parallel branch

Calculate voltage drop across the parallel resistors:
$$V_{parallel} = I_d \times R_{parallel} = \frac{99}{340}A \times \frac{120}{11}\Omega = \frac{108}{34} \approx 3.176V$$
This is $V_a = V_b = V_c = 3.176V$

Step5: Find currents for a, b, c

Use Ohm's Law for each parallel resistor:
$$I_a = \frac{V_a}{R_a} = \frac{\frac{108}{34}V}{30\Omega} = \frac{108}{1020} \approx 0.106A$$
$$I_b = I_a \approx 0.106A \quad (\text{same } R \text{ as a})$$
$$I_c = \frac{V_c}{R_c} = \frac{\frac{108}{34}V}{40\Omega} = \frac{108}{1360} \approx 0.079A$$

Step6: Calculate power for each resistor

Use $P = V \times I$ for each component:
$$P_a = V_a \times I_a = \frac{108}{34}V \times \frac{108}{1020}A = \frac{11664}{34680} \approx 0.336W$$
$$P_b = P_a \approx 0.336W$$
$$P_c = V_c \times I_c = \frac{108}{34}V \times \frac{108}{1360}A = \frac{11664}{46240} \approx 0.252W$$
$$V_d = 9V - V_{parallel} = 9V - \frac{108}{34}V = \frac{306 - 108}{34}V = \frac{198}{34} \approx 5.824V$$
$$P_d = V_d \times I_d = \frac{198}{34}V \times \frac{99}{340}A = \frac{19602}{11560} \approx 1.696W$$

Answer:

Source	a	b	c	d
$I\ (A)$	$\approx0.106$	$\approx0.106$	$\approx0.079$	$\approx0.291$
$V\ (V)$	$\approx3.18$	$\approx3.18$	$\approx3.18$	$\approx5.82$
$P\ (W)$	$\approx0.336$	$\approx0.336$	$\approx0.252$	$\approx1.696$