Circuit Physics Example 2

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Example 2

medium
A circuit has a 12 V12 \text{ V} battery, a 4 Ω4 \text{ } \Omega resistor and a 6 Ω6 \text{ } \Omega resistor in series, with a switch. When the switch is closed, what is the current and the voltage across each resistor?

Solution

  1. 1
    Total resistance in series: RT=4+6=10 ΩR_T = 4 + 6 = 10 \text{ } \Omega.
  2. 2
    Current (same throughout series circuit): I=VRT=1210=1.2 AI = \frac{V}{R_T} = \frac{12}{10} = 1.2 \text{ A}.
  3. 3
    Voltage across 4 Ω4 \text{ } \Omega: V1=IR1=1.2×4=4.8 VV_1 = IR_1 = 1.2 \times 4 = 4.8 \text{ V}. Voltage across 6 Ω6 \text{ } \Omega: V2=IR2=1.2×6=7.2 VV_2 = IR_2 = 1.2 \times 6 = 7.2 \text{ V}. Check: 4.8+7.2=12 V4.8 + 7.2 = 12 \text{ V}.

Answer

I=1.2 A;V1=4.8 V,V2=7.2 VI = 1.2 \text{ A}; \quad V_1 = 4.8 \text{ V}, \quad V_2 = 7.2 \text{ V}
In a series circuit, the same current flows through all components. The battery voltage divides among the resistors in proportion to their resistances, and the sum of voltage drops equals the battery voltage.

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