Correct option is A
The correct answer is:(a) 40V.
To find:
The voltageVVVacross the entire series circuit.
The three resistors are connected in series:R1=2ΩR_1 = 2 \, \OmegaR1=2Ω,R2=3ΩR_2 = 3 \, \OmegaR2=3Ω, andR3=5ΩR_3 = 5 \, \OmegaR3=5Ω.
The current through the 2Ω resistor isI=4AI = 4 \, \text{A}I=4A.
The resistors are connected to a battery with voltageVVV.
Calculate the total resistance in the series circuit:Since the resistors are connected in series, the total resistanceRtotalR_{\text{total}}Rtotalis the sum of the individual resistances:
Rtotal=R1+R2+R3=2Ω+3Ω+5Ω=10ΩR_{\text{total}} = R_1 + R_2 + R_3 = 2 \, \Omega + 3 \, \Omega + 5 \, \Omega = 10 \, \OmegaRtotal=R1+R2+R3=2Ω+3Ω+5Ω=10Ω
Use Ohm's Law to find the voltageVV
V:Ohm's Law states that:
Substituting the known values:
V=I⋅RtotalV = I \cdot R_{\text{total}}V=I⋅RtotalR_{total}Rtotal
V=4A×10Ω=40VV = 4 \, \text{A} \times 10 \, \Omega = 40 \, \text{V}V=4A×10Ω=40V
Conclusion:
The voltage across the series circuit isV=40VV = 40 \, \text{V}V=40V.
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