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In a series circuit with a 12V battery and three resistors of 2Ω, 3Ω, and 5Ω, wh
In a series circuit with a 12V battery and three resistors of 2Ω, 3Ω, and 5Ω, what is the current flowing through the circuit?
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Q1
In a series circuit with a 12V battery and three resistors of 2Ω, 3Ω, and 5Ω, what is the current flowing through the circuit?
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4A
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Total resistance R = 2Ω + 3Ω + 5Ω = 10Ω. Current I = V / R = 12V / 10Ω = 1.2A.
Questions & Step-by-step Solutions
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Q
Q: In a series circuit with a 12V battery and three resistors of 2Ω, 3Ω, and 5Ω, what is the current flowing through the circuit?
Solution:
Total resistance R = 2Ω + 3Ω + 5Ω = 10Ω. Current I = V / R = 12V / 10Ω = 1.2A.
Steps: 7
Show Steps
Step 1: Identify the voltage of the battery, which is 12V.
Step 2: Identify the resistors in the circuit, which are 2Ω, 3Ω, and 5Ω.
Step 3: Calculate the total resistance in the circuit by adding the resistances together: 2Ω + 3Ω + 5Ω.
Step 4: Perform the addition: 2 + 3 = 5, then 5 + 5 = 10. So, the total resistance R is 10Ω.
Step 5: Use Ohm's Law to find the current. Ohm's Law states that Current (I) = Voltage (V) / Resistance (R).
Step 6: Substitute the values into the formula: I = 12V / 10Ω.
Step 7: Perform the division: 12 divided by 10 equals 1.2. So, the current I is 1.2A.
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