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Answer :
Sure! Let's break down the solution to the problem about the series circuit with a 9 V battery, a 3 Ω resistor, and a 6 Ω resistor:
1. Find the Total Resistance:
- In a series circuit, the total resistance is simply the sum of all individual resistances. So, you add the resistances together:
[tex]\[
\text{Total Resistance} = 3 \, \Omega + 6 \, \Omega = 9 \, \Omega
\][/tex]
2. Find the Current:
- Use Ohm's Law to determine the current flowing through the circuit. 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.
- Rearrange the formula to solve for current ([tex]\( I \)[/tex]):
[tex]\[
I = \frac{V}{R} = \frac{9 \, \text{V}}{9 \, \Omega} = 1 \, \text{A}
\][/tex]
- So, the current flowing through the circuit is 1 ampere (A).
3. Find the Voltage Drop Across Each Resistor:
- To find the voltage drop across each resistor, use Ohm's Law again. The voltage drop ([tex]\( V \)[/tex]) across a resistor in a series circuit is given by [tex]\( V = I \times R \)[/tex].
- Voltage Drop Across the 3 Ω Resistor:
[tex]\[
V = 1 \, \text{A} \times 3 \, \Omega = 3 \, \text{V}
\][/tex]
- Voltage Drop Across the 6 Ω Resistor:
[tex]\[
V = 1 \, \text{A} \times 6 \, \Omega = 6 \, \text{V}
\][/tex]
So, in conclusion:
- The total resistance in the circuit is 9 Ω.
- The current flowing through the circuit is 1 A.
- The voltage drop across the 3 Ω resistor is 3 V.
- The voltage drop across the 6 Ω resistor is 6 V.
1. Find the Total Resistance:
- In a series circuit, the total resistance is simply the sum of all individual resistances. So, you add the resistances together:
[tex]\[
\text{Total Resistance} = 3 \, \Omega + 6 \, \Omega = 9 \, \Omega
\][/tex]
2. Find the Current:
- Use Ohm's Law to determine the current flowing through the circuit. 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.
- Rearrange the formula to solve for current ([tex]\( I \)[/tex]):
[tex]\[
I = \frac{V}{R} = \frac{9 \, \text{V}}{9 \, \Omega} = 1 \, \text{A}
\][/tex]
- So, the current flowing through the circuit is 1 ampere (A).
3. Find the Voltage Drop Across Each Resistor:
- To find the voltage drop across each resistor, use Ohm's Law again. The voltage drop ([tex]\( V \)[/tex]) across a resistor in a series circuit is given by [tex]\( V = I \times R \)[/tex].
- Voltage Drop Across the 3 Ω Resistor:
[tex]\[
V = 1 \, \text{A} \times 3 \, \Omega = 3 \, \text{V}
\][/tex]
- Voltage Drop Across the 6 Ω Resistor:
[tex]\[
V = 1 \, \text{A} \times 6 \, \Omega = 6 \, \text{V}
\][/tex]
So, in conclusion:
- The total resistance in the circuit is 9 Ω.
- The current flowing through the circuit is 1 A.
- The voltage drop across the 3 Ω resistor is 3 V.
- The voltage drop across the 6 Ω resistor is 6 V.
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