Answer to Question #168570 in Electric Circuits for Chaihcai

Question

Given: Resistor 1= 5 ohm, Resistor 2= 12 ohm, resistor 3= 6 ohm, resistor 4= 6 ohm and resistor 5= 30 ohm.

1. Suppose the four resistors are connected in parallel. What is the resistance of the combined resistors?

2. Suppose the four resistors are connected in parallel. Find (a) equivalent resistance (b) current flowing through each resistor (c) voltage across each resistor and (d) power dissipated by each resistor.

3. Find the equivalent resistance of the combination of resistance shown.

Answer 1

1) The equivalent resistance of the resistors connected in parallel can be found as follows:

$$\dfrac{1}{R_{eq}}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}+\dfrac{1}{R_{4}},$$ $$R_{eq}=\dfrac{1}{\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}+\dfrac{1}{R_{4}}},$$ $$R_{eq}=\dfrac{1}{\dfrac{1}{5\ \Omega}+\dfrac{1}{12\ \Omega}+\dfrac{1}{6\ \Omega}+\dfrac{1}{6\ \Omega}}=1.62\ \Omega.$$

2) (a) The equivalent resistance of the resistors connected in parallel can be found as follows:

$$\dfrac{1}{R_{eq}}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}+\dfrac{1}{R_{4}},$$ $$R_{eq}=\dfrac{1}{\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}+\dfrac{1}{R_{4}}},$$ $$R_{eq}=\dfrac{1}{\dfrac{1}{5\ \Omega}+\dfrac{1}{12\ \Omega}+\dfrac{1}{6\ \Omega}+\dfrac{1}{6\ \Omega}}=1.62\ \Omega.$$

(b)-(c) Unfortunately, from the definition of the question we don't know neither the voltage supply connected across the parallel combination of the resistors nor the total current drawn by the combination of the resistors. Let's suppose that the 12-V voltage supply connected across the parallel combination of the resistors. In the parallel ombination of resistors, the voltage across each of the resistors is the same, so we can write:

$$V=V_1=V_2=V_3=V_4=12\ V.$$

Finally, from the Ohm's Law we can find the current flowing through each resistor:

$$I_1=\dfrac{V}{R_1}=\dfrac{12\ V}{5\ \Omega}=2.4\ A,$$ $$I_2=\dfrac{V}{R_2}=\dfrac{12\ V}{12\ \Omega}=1\ A,$$ $$I_3=\dfrac{V}{R_3}=\dfrac{12\ V}{6\ \Omega}=2\ A,$$ $$I_4=\dfrac{V}{R_1}=\dfrac{12\ V}{6\ \Omega}=2\ A.$$

(d) We can find the power dissipated by each resistor as follows:

$$P_1=I_1^2R_1=(2.4\ A)^2\cdot5\ \Omega=28.8\ W,$$ $$P_2=I_2^2R_2=(1\ A)^2\cdot12\ \Omega=12\ W,$$ $$P_3=I_3^2R_3=(2\ A)^2\cdot6\ \Omega=24\ W,$$ $$P_4=I_4^2R_4=(2\ A)^2\cdot6\ \Omega=24\ W.$$