Question

Parallel Circuits

Master Page: https://questioncove.com/study#/updates/5ae2c3a7293a9d6f092ba200

Answer 1

Parallel circuits are noted as being circuits with multiple pathways. Voltage is constant across each branch because they each have access to the battery. The current is not constant, unlike series circuits.

Answer 2

^Circuits for problems #1 - 4

Answer 3

Firstly, note that in all problems (c) where you are solving for the total current provided by the battery, you may notice the following relationship
\[I_{total} = I_{1} + I_{2} + I{...}\]
The total current is equal to the sum of the values of each of the currents flowing through the branches. If you notice this relationship and are confident in applying it, it is much quicker than having to solve for the total resistance first.

I do not solve for the current using the above method to demonstrate the use of the formula and an easy way to find the total resistance. If you can remember the concept though, go for it.

Secondly, when solving for the total resistance, I demonstrate it as the following:
\[R_{total} = R_{1}^{-1} + R_{2}^{-1} + R_{....}^{-1} = x \Omega\]
Keep in mind that once you sum the inverses of each resistance, you must take the inverse of that sum in order to find the correct total resistance. For example:
\[R_{total} = 2^{-1} + 4^{-1} = 0.75\]
We must then inverse 0.75 to get:
\[R_{total} = 1.33 \Omega\]
I skip this step in my solving below. If you get a number that doesn't match mine for the total resistance, be sure you are taking the inverse of the sum.

Answer 4

\({\bf{Practice~1}}\)
1. Use the parallel circuit to solve questions (a) - (d).
a. What is the voltage across each bulb?
\[V_{2} = V_{2} = V_{total} = 12 V\]
b. What is the current in each branch?
\[I_{2} = \frac{ V_{2} }{ R_{2} }\]
\[I_{2} = \frac{ 12 }{ 2} = 6A\]
c. What is the total current provided by the battery?
\[I_{total} = \frac{ V_{total} }{ R_{total} }\]
\[R_{total} = 2^{-1} + 2^{-1} = 1 \Omega\]
\[I_{total} = \frac{ 12 }{ 1 } = 12 A\]
d. Use the total current and the total voltage to calculate the total resistance.
\[R_{total} = \frac{ V_{total} }{ I_{total} }\]
\[R_{total} = \frac{ 12 }{ 12 } = 1 \Omega\]
2. Use the parallel circuit to answer questions (a) - (d).
a. What is the voltage across each bulb?
\[V_{3} = V_{3} = V_{total} = 12 V\]
b. What is the current in each branch?
\[I_{3} = \frac{ V_{3} }{ R_{3} }\]
\[I_{3} = \frac{ 12 }{ 3 } = 4 A\]
c. What is the total current provided by the battery?
\[I_{total} = \frac{ V_{total} }{ R_{total} } \]
\[R_{total} = 3^{-1} + 3^{-1} = 1.5 \Omega\]
\[I_{total} = \frac{ 12 }{ 1.5 } = 8 A\]
d. Use the total current and the total voltage to calculate the total resistance of the circuit?
\[R_{total} = \frac{ V_{total} }{ I_{total} }\]
\[R_{total} = \frac{ 12 }{ 8 } = 1.5 \Omega\]
3. Use the parallel circuit to answer questions (a) - (d).
a. What is the voltage across each resistor?
\[V_{2} = V_{3} = V_{total} = 12 V\]
b. What is the current in each branch?
\[I_{2} = \frac{ V_{2} }{ R_{2}}\]
\[I_{2} = \frac{ 12 }{ 2} = 6 A\]
\[I_{3} = \frac{ V_{3} }{ R_{3}}\]
\[I_{3} = \frac{ 12 }{ 3 } = 4 A\]
c. What is the total current provided by the batteries?
\[I_{total} = \frac{ V_{total}}{ R_{total} }\]
\[R_{total}= 2^{-1} + 3^{-1} = 1.2 \Omega\]
\[I_{total} = \frac{ 12 }{ 1.2 } = 10 A\]
d. Use the total current and the total voltage to calculate the total resistance of the circuit.
\[R_{total} = \frac{ V_{total} }{ I_{total} }\]
\[R_{total} = \frac{ 12 }{ 10 } = 1.2 \Omega\]
4. Use the parallel circuit to answer questions (a) - (c)
a. What is the voltage across each resistor?
\[V_{2} = V_{3} = V_{1} = V_{total} = 9 V\]
b. What is the current in each branch?
\[I_{2} = \frac{ V_{2} }{ R_{2} }\]
\[I_{2} = \frac{ 9 }{ 2 } = 4.5 A\]
\[I_{3} = \frac{ V_{3} }{ R_{3} }\]
\[I_{3} = \frac{ 9 }{ 3 } = 3 A\]
\[I_{1} = \frac{ V_{1} }{ R_{1} }\]
\[I_{1} = \frac{ 9 }{ 1 } = 9 A\]
c. What is the total current provided by the battery?
\[I_{total} = \frac{ V_{total} }{ R_{total} }\]
\[R_{total} = 2^{-1} + 3^{-1} + 1^{-1} = 0.54 \Omega\]
\[I_{total} = \frac{ 9 }{ 0.54 } = 16.5 A\]

Answer 5

\({\bf{Practice~2}}\)
1. Calculate the total resistance of a circuit containing each of the following combinations of resistors
a. Two 8 ohm resistors in parallel
\[R_{total} = 8^{-1} + 8^{-1} = 0.25^{-1} = 4 \Omega\]
b. Two 12 ohm resistors in parallel
\[R_{total} = 12^{-1} + 12^{-1} = 0.166^{-1} = 6 \Omega\]
c. A 4 ohm resistor and an 8 ohm resistor in parallel
\[R_{total} = 4^{-1} + 8^{-1} = 0.375^{-1} = 2.7 \Omega\]
d. A 12 ohm resistor and a 3 ohm resistor in parallel
\[R_{total} = 12^{-1} + 3^{-1} = 0.416^{-1} = 2.4 \Omega\]
2. To find the total resistance of three resistors A, B, and C in parallel, first use the formula to find the total of resistors A and B. Then use the formula again to combine resistor C with the total of a and B. use this method to find the total resistance of a circuit containing each of the following combinations of resistors.
a. Three 8 ohm resistors in parallel
\[R_{total} = 8^{-1} + 8^{-1} + 8^{-1} = 0.375^{-1} = 2.7 \Omega\]
b. Two 6 ohm resistors and a 2 ohm resistor in parallel
\[R_{total} = 6^{-1} + 6^{-1} + 2^{-1} = 0.833^{-1} = 1.2 \Omega\]
c. A 1 ohm resistor, a 2 ohm resistor, and a 3 ohm resistor in parallel
\[R_{total} = 1^{-1} + 2^{-1} + 3^{-1} = 1.833^{-1} = 0.54 \Omega\]