# Experts please answer... Question 1 Draw a circuit diagram showing a cell, a bulb and a closed switch. Question 2 ??????????????a)??How much work is done in moving a charge of 3 coulumb from a point at the volts 115 to a point at 125 volts? ??????????????b)??Ammeter burns out when connected in parallel. Give reasons. Question 3 Given n resistors each of resistors Rs. How will you combine them to get the (i) maximum and (ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance? Question 4 a) A wire of length L and resistance R is stretched so that its length it?s doubled. How will the: (a) Resistance change(b) Resistivity change? ??????????????b)??In an experiment the current flowing through a resistor and potential difference across it are measured. The values are given below. Show that these values confirm Ohm?s Law ??????????????I (ampere) 1.0 1.0 2 1.5 2.0 2.0 2.5 2.5 3.0 3.0 ??????????????V (volt) 4.0 4.0 6.0 6.0 8.0 8.0 10.0 10.0 12.0 12.0 Question 5 a) A tube light draws 0.1 A current from a 220 V supply. What current will this tube light draw when it is connected to a 110 V supply? b) An electric wire is stretched to increase its length by 25%.By what % will the resistance be increased and what will be increase in its resistivity? Thank you so much! :) For remaining queries we request you to post them in separate threads to have rapid assistance from our experts.
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Here is a better pic of the questions • 0
2) a) we know that
charge = q = 3 coulombs
potential difference = 125 - 115 =10v
Energy =?
FORMULA: POTENTIAL DIFFERENCE = WORK DONE ÷ CHARGE
V = W ÷ Q
10 = W ÷ 3
W = 3 × 10
W = 30JOULES

Therefore, The work done in moving a charge of 3 coulomb from a point at volts 115 to a point at 125 volts is 30 JOULES
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2) b) An ammeter is likely to burn if connected in parallel because ammeter is a low resistance device and when connected in parallel, the resistance of the circuit reduces considerably. Hence, a large current flows in the circuit which may lead to burning of a circuit.
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5) a)

I =0.1

V=220

applying ohms law

R=V/I

R=2200

putting value of R in

V=110

R=2200

I= V/R

I= 110/2200

I=0.05

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b)
Increase in length by 25%:
125%/100% × L = 1.25 L
Resistance is usually given by the formula :
The dimensions for the initial wire are:
R = kL/A
Length is L and a cross-sectional area of A
new wire will have dimension of Length 1.25L and Cross-sectional area of A.
Calculate the resistance of the wire at first and after being stretched:
Initial resistance = kL/A
Final resistance = k1.25L/A

Find % increase = 1.25Lk/A ÷ kL/A × 100%
= 1.25kL/A × A/kL ×100%
=125%
125 % - 100% = 25%
There will be a 25 % increase in resistance.
Resistance is dependent on three factors:
1) Material from which material is made from. This is it's resistivity (k)and is a constant. It does not change in the same material
2) Length
3) Cross sectional area.

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