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Questions 39-41 refer to the following figure:FIGURE CANNOT COPYIf $V=100 \mathrm{V}, R_{1}=50 \Omega, R_{2}=80 \Omega$ and $R_{3}=120 \Omega,$ determine the voltage across $R_{3}$ .(A) 100 V(B) 60 V(C) 40 V(D) 20 V

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Chapter 12

Practice Test 2

Section 1

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Hi in the given problem. The two resistors R two and R three first of all are in series then. But resistor R one is in parallel with these two. This combination is connected, yep. A source of voltage providing voltage. V The value of this week is 100 world resistance are one is 50 oh R two is mhm. Be home and our three is 120 no. As voltage remains seen in parallel. So the series combination of R. one and R two will get seen. We is equal 200 walls across them. Therefore current passing through them will be given by Homes Law as I used to be by our for our disease are too plus our three the net resistance in serious combinations. So here it will be underworld divided by 80 Last 120 mm. Yes this is hunted by 200 or we can say current passing through the city's combination A 0.5 and here now as current remains the same in series. So the current across this artery will also be seen 0.5 Ampere. Hence voltage drop a cruise. Our three will be using arms law again. That will be given by I into our This is I into our three For either 0.5 and pure multiplied by our three 120 home. So finally this voltage here comes out to be 60 world. Hence we can say, yeah, our option B is correct. Yeah. Thank you.

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