Question

You wire three resistors in a circuit; a 110Ω, a 220Ω and a 440Ω across a 9.0V battery.  Determine the current in the circuit, the current in each resistor and the voltage drop in each resistor when:

1. The resistors are wired in series.
2. The resistors are wired in parallel.

Answer 1

1) Let's first find the equivalent resistance of the resistors connected in series:

$$R_{eq}=R_1+R_2+R_3,$$ $$R_{eq}=110\ \Omega+220\ \Omega+440\ \Omega=770\ \Omega.$$

Then, we can find the current flowing through the equivalent resistance (or the current in the circuit) from the Ohm's law:

$$I=\dfrac{V}{R}=\dfrac{9\ V}{770\ \Omega}=0.012\ A.$$

Since the amount of current that flows through a resistors connected in series is the same, we can write:

$$I=I_1=I_2=I_3=0.012\ A.$$

Finally, we can find the voltage drop in each resistor:

$$V_1=I_1R_1=0.012\ A\cdot110\ \Omega=1.32\ V,$$ $$V_2=I_2R_2=0.012\ A\cdot220\ \Omega=2.64\ V,$$ $$V_3=I_3R_3=0.012\ A\cdot440\ \Omega=5.28\ V.$$

2) Let's first find the equivalent resistance of the resistors connected in parallel:

$$\dfrac{1}{R_{eq}}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}},$$ $$R_{eq}=\dfrac{1}{\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}}$$ $$R_{eq}=\dfrac{1}{\dfrac{1}{110\ \Omega}+\dfrac{1}{220\ \Omega}+\dfrac{1}{440\ \Omega}}=62.86\ \Omega.$$

Then, we can find the current flowing through the equivalent resistance (or the current in the circuit) from the Ohm's law:

$$I=\dfrac{V}{R}=\dfrac{9\ V}{62.86\ \Omega}=0.143\ A.$$

Since the voltage drop is the same across all the elements in the parallel circuit, we can write:

$$V=V_1=V_2=V_3=9\ V.$$

Finally, we can find the current in each resistor from the Ohm's law:

$$I_1=\dfrac{V}{R_1}=\dfrac{9\ V}{110\ \Omega}=0.082\ A,$$ $$I_2=\dfrac{V}{R_2}=\dfrac{9\ V}{220\ \Omega}=0.041\ A,$$ $$I_3=\dfrac{V}{R_3}=\dfrac{9\ V}{440\ \Omega}=0.02\ A.$$