Answer to Question #170081 in Electric Circuits for Adaora

Question

A resistance of 5ohms is connected in parallel with another of a 4ohm and the parallel arrangements is connected in series with a 2ohm resistor and an end of 20V. Calculate the potential difference and current across each resistor

Answer 1

Let's first find the equivalent resistance of the parallel arrangements of resistors:

$$\dfrac{1}{R_{eq,p}}=\dfrac{1}{R_1}+\dfrac{1}{R_2},$$ $$R_{eq,p}=\dfrac{R_1R_2}{R_1+R_2}=\dfrac{5\ \Omega\cdot4\ \Omega}{5\ \Omega+4\ \Omega}=2.22\ \Omega.$$

Let's find the equivalent resistance of this parallel arrangements of resistors and the resistor connected in series:

$$R_{eq, s}=R_{eq, p}+R_3=2.22\ \Omega+2\ \Omega=4.22\ \Omega.$$

Then, we can find the current in the circuit from the Ohm's law:

$$I=\dfrac{V}{R_{eq,s}}=\dfrac{20\ V}{4.22\ \Omega}=4.74\ A.$$

Since the current in the series circuit is the same, $I=I_3=I_{eq,p}=4.74\ A.$

Then, we can find the potential difference on the resistor $R_3$:

$$V_3=I_3R_3=2\ \Omega\cdot4.74\ A=9.48\ V.$$

Then, we can find the potential difference on the resitor $R_{eq, p}$:

$$V_{eq,p}=IR_{eq,p}=4.74\ A\cdot2.22\ \Omega=10.52\ V.$$

In the parallel circuits the voltage is the same across all elements, therefore:

$$V_{eq,p}=V_1=V_2=10.52\ V.$$

Then, we can find the currents across resistors $R_1$ and $R_2$ from the Ohm's law:

$$I_1=\dfrac{V_1}{R_1}=\dfrac{10.52\ V}{5\ \Omega}=2.1\ A,$$ $$I_2=\dfrac{V_2}{R_2}=\dfrac{10.52\ V}{4\ \Omega}=2.63\ A.$$

Answer:

$V_1=V_2=10.52\ V, V_3=9.48\ V.$

$I_1=2.1\ A, I_2=2.63\ A, I_3=4.74\ A.$