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In a series circuit with a 12V battery and two resistors of 3Ω and 6Ω, what is t
In a series circuit with a 12V battery and two resistors of 3Ω and 6Ω, what is the current flowing through the circuit?
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Q1
In a series circuit with a 12V battery and two resistors of 3Ω and 6Ω, what is the current flowing through the circuit?
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Total resistance R = R1 + R2 = 3Ω + 6Ω = 9Ω. Current I = V/R = 12V/9Ω = 4/3A.
Questions & Step-by-step Solutions
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Q
Q: In a series circuit with a 12V battery and two resistors of 3Ω and 6Ω, what is the current flowing through the circuit?
Solution:
Total resistance R = R1 + R2 = 3Ω + 6Ω = 9Ω. Current I = V/R = 12V/9Ω = 4/3A.
Steps: 6
Show Steps
Step 1: Identify the voltage of the battery, which is 12V.
Step 2: Identify the resistors in the circuit, which are 3Ω and 6Ω.
Step 3: Calculate the total resistance in the circuit by adding the resistances together: 3Ω + 6Ω = 9Ω.
Step 4: Use Ohm's Law to find the current. Ohm's Law states that Current (I) = Voltage (V) / Resistance (R).
Step 5: Substitute the values into the formula: I = 12V / 9Ω.
Step 6: Simplify the calculation: 12V / 9Ω = 4/3A.
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