Answer :
To find the voltage of the battery in a series circuit with given resistors and current, we can use Ohm’s Law and the properties of series circuits. Here's how you can solve it:
1. Understand the problem: You have a series circuit with three resistors each of [tex]\(2 \, \Omega\)[/tex] and one resistor of [tex]\(6 \, \Omega\)[/tex]. The current flowing through each resistor is [tex]\(4.0 \, \text{A}\)[/tex].
2. Total Resistance in Series: In a series circuit, the total resistance is the sum of the individual resistances. So, you add up all the resistors:
[tex]\[
R_{\text{total}} = 2 \, \Omega + 2 \, \Omega + 2 \, \Omega + 6 \, \Omega = 12 \, \Omega
\][/tex]
3. Apply Ohm’s Law: Ohm’s Law states that [tex]\(V = I \times R\)[/tex], where [tex]\(V\)[/tex] is voltage, [tex]\(I\)[/tex] is current, and [tex]\(R\)[/tex] is resistance.
4. Calculate the Voltage: Using the total resistance and the current, calculate the voltage:
[tex]\[
V = I \times R_{\text{total}} = 4.0 \, \text{A} \times 12 \, \Omega = 48 \, \text{V}
\][/tex]
So, the voltage of the battery in the circuit is [tex]\(48 \, \text{V}\)[/tex].
1. Understand the problem: You have a series circuit with three resistors each of [tex]\(2 \, \Omega\)[/tex] and one resistor of [tex]\(6 \, \Omega\)[/tex]. The current flowing through each resistor is [tex]\(4.0 \, \text{A}\)[/tex].
2. Total Resistance in Series: In a series circuit, the total resistance is the sum of the individual resistances. So, you add up all the resistors:
[tex]\[
R_{\text{total}} = 2 \, \Omega + 2 \, \Omega + 2 \, \Omega + 6 \, \Omega = 12 \, \Omega
\][/tex]
3. Apply Ohm’s Law: Ohm’s Law states that [tex]\(V = I \times R\)[/tex], where [tex]\(V\)[/tex] is voltage, [tex]\(I\)[/tex] is current, and [tex]\(R\)[/tex] is resistance.
4. Calculate the Voltage: Using the total resistance and the current, calculate the voltage:
[tex]\[
V = I \times R_{\text{total}} = 4.0 \, \text{A} \times 12 \, \Omega = 48 \, \text{V}
\][/tex]
So, the voltage of the battery in the circuit is [tex]\(48 \, \text{V}\)[/tex].
