Question

Section A: Complete this section within 90 minutes of the examination start time. Answer on the MCQ card. NB these will be collected about 30 minutes before the end of the examination. Section B: Answer this section in the answer book provided. Show your working and include diagrams and units in all questions where relevant. SECTION A: MULTIPLE CHOICE; total marks = 40 Choose the option that best answers the question. 1. The equivalent capacitance (in ?F) of the capacitor network shown – i.e. between points a and b is: A. 1 B. 2 C. 3 D. 4 E. 5 [3] 2. What is the charge on the capacitor and how much energy is stored in the capacitor in the circuit shown? A. 81 ?C; 18 ?J B. 18 ?C; 81 ?J C. 18 ?C; 9 ?J D. 81 C; 81 J E. 18 J; 81 ?C [4] 3. A 20 V lamp designed to dissipate 80 W is connected in series with a resistance of R and a 60 V battery. What value must R be for the lamp to operate as rated? A. 5 ? B. 4 ? C. 6 ? D. 8 ? E. 10 ? [3]

          Section A: Complete this section within 90 minutes of the examination start time. Answer on the MCQ card. NB these will be collected about 30 minutes before the end of the examination.
Section B: Answer this section in the answer book provided. Show your working and include diagrams and units in all questions where relevant.
SECTION A: MULTIPLE CHOICE; total marks = 40
Choose the option that best answers the question.
1. The equivalent capacitance (in ?F) of the capacitor network shown – i.e. between points a and b is:
A. 1
B. 2
C. 3
D. 4
E. 5
[3]
2. What is the charge on the capacitor and how much energy is stored in the capacitor in the circuit shown?
A. 81 ?C; 18 ?J
B. 18 ?C; 81 ?J
C. 18 ?C; 9 ?J
D. 81 C; 81 J
E. 18 J; 81 ?C
[4]
3. A 20 V lamp designed to dissipate 80 W is connected in series with a resistance of R and a 60 V battery. What value must R be for the lamp to operate as rated?
A. 5 ?
B. 4 ?
C. 6 ?
D. 8 ?
E. 10 ?
[3]

Show more…

Section A: Complete this section within 90 minutes of the examination start time. Answer on the MCQ card. NB these will be collected about 30 minutes before the end of the examination.
Section B: Answer this section in the answer book provided. Show your working and include diagrams and units in all questions where relevant.
SECTION A: MULTIPLE CHOICE; total marks = 40
Choose the option that best answers the question.
1. The equivalent capacitance (in ?F) of the capacitor network shown – i.e. between points a and b is:
A. 1
B. 2
C. 3
D. 4
E. 5
[3]
2. What is the charge on the capacitor and how much energy is stored in the capacitor in the circuit shown?
A. 81 ?C; 18 ?J
B. 18 ?C; 81 ?J
C. 18 ?C; 9 ?J
D. 81 C; 81 J
E. 18 J; 81 ?C
[4]
3. A 20 V lamp designed to dissipate 80 W is connected in series with a resistance of R and a 60 V battery. What value must R be for the lamp to operate as rated?
A. 5 ?
B. 4 ?
C. 6 ?
D. 8 ?
E. 10 ?
[3]

Added by Lisa B.

University Physics with Modern Physics

Hugh D. Young 13th Edition

Chapter 26

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Solved by Expert Vishal Gupta

11/08/2023

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Transcript

00:01 Hi, here in this given problem there are three different questions based upon current electricity.

00:06 The first one it is based upon the combination of capacitors, the mixed grouping series and parallel.

00:16 So that is something like this.

00:37 The terminals a and b, all these capacitors are of 2 microfarad each.

00:46 This is also 2 microfarad and these, the common capacitors, these are 1 microfarad each.

01:02 Now to find the net capacitance starting with this series combination, we know in series if identical capacitors are joined in series, the net capacitance is given as the value of one capacitor divided by the number of capacitors as these are two only.

01:19 So it comes out to be equal to 1 microfarad.

01:22 Then this one is in parallel with this one.

01:25 So cp, that is 1 plus 1 means 2 microfarad.

01:29 Then again, as it becomes 2 microfarad, so again it is in series with this two.

01:34 So again it will become 2 and 2, 1 microfarad.

01:38 Then 1 plus 1 again 2.

01:41 Then this 2 and 2 again in series 1.

01:44 1 and 1 again in parallel.

01:46 So 2, then again 2 and 2 in series.

01:49 So becomes 1 and 1 and 1 again in parallel becomes 2.

01:53 So finally 2 and 2 in series.

01:57 So net capacitance of this combination comes out to be equal to 1 microfarad only.

02:04 And this is the option a which is correct over here.

02:09 Now in the second problem.

02:12 This is the circuit having a resistor resistance 3 ohm.

02:19 In parallel with it, there is a capacitor.

02:23 Capacitor having capacitance 2 microfarad.

02:27 Then a cell providing 12 volt and its internal resistance 1 ohm.

02:35 As we know in a steady state, there will be no current through capacitor...

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