In the given circuit the ammeter A 1 and A 2 are ideal and the ammeter A 3 has a resistance of 1 . 9 × 10 – 3 Ω . If sum of readings of all three meters is given by 2 n 27 Ampere the value of n will be.

Let the currents I , I 1 , I 2 be flowing in the networks as shown above in the diagram, Applying KVL in the loops,. For loop A O B A , 3 I - 8 + 5 ( I - I 2 ) - 10 = 0 ⇒ 8 I - 5 I 2 = 18 . . . ( 1 ) For loop A O C A , 5 ( I - I 2 ) - 10 + 20 - 15 I 2 = 0 ⇒ I = 4 I 2 - 2 . . . ( 2 ) , For loop B O C B , 3 I - 8 + 1 . 9 × 10 - 3 I 1 + 15 I 2 - 20 = 0 ⇒ 3 I + 15 I 2 + 1 . 9 × 10 - 3 I 1 = 28 . . . ( 1 ) From equations 1 , 2 , 3 , we get, I = 82 27 , I 1 = 0 , I 2 = 34 27 . These are the currents flowing in the given ammeters and also their sum is given as 2 n 27 , i.e., 2 n 27 = 82 27 + 34 27 + 0 ⇒ n = 116 2 = 58 . Hence, the value of n = 58 .

Two resistors, 400 Ω , and 800 Ω are connected in series with a 6 V battery. It is desired to measure the current in the circuit. An ammeter of 10 Ω resistance is used for this purpose. The reading of ammeter will be N 1210 A . Similarly, if a voltmeter of 1000 Ω resistance is used to measure the potential difference across the 400 Ω resistor, the reading of voltmeter is P 19 V . Then the value of N and P are:

Diagram for the first case when ammeter is connected to measure current is shown below. Since, ammeter has a resistance of 10 Ω and is connected in series with the other two resistors. So, the equivalent resistance in the circuit will be, R eq = 400 + 800 + 10 = 1210 Ω , Given supply voltage is, V = 6 V , So, current in the circuit, I is; I = V R = 6 1210 A , Now, comparing the calculated current with the ammeter reading, we get N = 6 . Again, the circuit diagram to measure voltage across the given resistor is shown below; Since, voltmeter has a resistance of 1000 Ω and is parallel with 400 Ω resistor and their combination is in series with 800 Ω resistor, then, The equivalent resistance in the circuit is given as; R eq = 800 + ( 1000 × 400 1000 + 400 ) = 800 + 2000 7 ⇒ R eq = 7600 7 Ω Now, current in the circuit is; I = V R eq = 6 7600 7 = 42 7600 A , Then, potential difference across resistor 400 Ω is; V = I R = 42 7600 × 2000 7 = 30 19 Now, comparing the calculated voltage and given reading we get, P = 30 .

The cross section area and length of a cylindrical conductor are A and l respectively and it is connected with a voltage source V 0 . The conductivity varies as σ = σ 0 l x where x ( 0 < x < l ) is the distance along the axis of the cylinder from one of its ends as shown in the figure. Choose the correct option.

The electric field in the wire at x in cylinder will be 2 V 0 l 2 x .

The electric current in the wire will be V 0 σ 0 A 2 l .

N cells each of e.m.f. E and identical resistance r are grouped into sets of K cells connected in series. The ( N K ) sets are connected in parallel to a load of resistance R , then;

Maximum power delivered to the load is N E 2 4 r .

Maximum power is delivered to the load if K = r N R .

Maximum power delivered to the load is E 2 4 N r .

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In the circuit shown, reading of the voltmeter connected across $400 Ω$ resistance is $60 V .$ If it is connected across $300 Ω$ resistance then reading in volt will be

Important Questions on Current Electricity

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Alpha Question Bank for Engineering: Physics>Chapter 24 - Current Electricity>Exercise - 2>Q 55

In the given circuit the ammeter $A_{1} and A_{2}$ are ideal and the ammeter $A_{3}$ has a resistance of $1.9 \times 10^{- 3} Ω$ . If sum of readings of all three meters is given by $(\frac{2 n}{27})$ Ampere the value of $n$ will be.

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Alpha Question Bank for Engineering: Physics>Chapter 24 - Current Electricity>Exercise - 2>Q 56

Two resistors,

400 Ω,

and

800 Ω

are connected in series with a

6 V

battery. It is desired to measure the current in the circuit. An ammeter of

10 Ω

resistance is used for this purpose. The reading of ammeter will be

\frac{N}{1210} A .

Similarly, if a voltmeter of

1000 Ω

resistance is used to measure the potential difference across the

400 Ω

resistor, the reading of voltmeter is

\frac{P}{19} V .

Then the value of

N

and

P

are:

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Alpha Question Bank for Engineering: Physics>Chapter 24 - Current Electricity>Exercise - 2>Q 57

A current passes through a wire of non-uniform cross-section. Which of the following quantities are independent of the cross-section?

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Alpha Question Bank for Engineering: Physics>Chapter 24 - Current Electricity>Exercise - 2>Q 58

When no current is passed through a conductor

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Alpha Question Bank for Engineering: Physics>Chapter 24 - Current Electricity>Exercise - 2>Q 59

The current density in a wire is

10 A {cm}^{- 2}

and the electric field in the wire is

5 V {cm}^{- 1} .

ρ =

resistivity of material,

σ =

conductivity of the material then (in

S . I .

units) :

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Alpha Question Bank for Engineering: Physics>Chapter 24 - Current Electricity>Exercise - 2>Q 60

A bulb is connected to an ideal battery of emf

10 V

so that the resulting current is

10 mA .

When the bulb is connected to

220 V

mains (ideal), the current is

50 mA .

Choose the correct alternative(s).

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Alpha Question Bank for Engineering: Physics>Chapter 24 - Current Electricity>Exercise - 2>Q 61

The cross section area and length of a cylindrical conductor are $A$ and $l$ respectively and it is connected with a voltage source $V_{0}$ . The conductivity varies as $σ = σ_{0} \frac{l}{x}$ where $x (0 < x < l)$ is the distance along the axis of the cylinder from one of its ends as shown in the figure. Choose the correct option.

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Alpha Question Bank for Engineering: Physics>Chapter 24 - Current Electricity>Exercise - 2>Q 62

N

cells each of e.m.f.

E

and identical resistance

r

are grouped into sets of

K

cells connected in series. The

(\frac{N}{K})

sets are connected in parallel to a load of resistance

R

, then;

In the circuit shown, reading of the voltmeter connected across 400 Ω resistance is 60 V. If it is connected across 300 Ω resistance then reading in volt will be

Important Questions on Current Electricity

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In the circuit shown, reading of the voltmeter connected across $400 Ω$ resistance is $60 V .$ If it is connected across $300 Ω$ resistance then reading in volt will be