Current Electricity Test - 10

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Current Electricity Test - 10

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Weekly Quiz Competition

Question 1
4 / -1

Point out the right statements about the validity of Kirchhoff's junction rule

A
it is based on conservation of charge

B
outgoing currents add up and are equal to incoming currents at a junction

C
bending or reorienting the wire does not change the validity of Kirchhoff's junction rule

D
all of above

Solution

By Kirchhoff's junction rule. Incoming current = Outgoing current
The net charge is conserved and it is based on conservation of charge.
Also bending or reorienting the wire does not invalidate the conservation of charge principle.
Question 2
4 / -1

The figure below shows currents in a part of electric circuit. The current i is

A
1.7 amp

B
3.7 amp

C
1.3 amp

D
1 amp

Solution

Applying Kirchhoffs first law,
1 = 2 + 2 - I - 1.3 = 1.7 A
Question 3
4 / -1

A circuit has a section ABC if the potential at point A, B and C are V₁, V₂and V₃ respectively, calculate the potential at point O is

A
V₁ + V₂ + V₃

B

C
zero

D

Solution

Applying juction rule −I₁− I₂ − I₃= 0
i.e., I₁+ I₂ + I₃ = 0
Let, V₀ bet the potential at point O. By Ohm's law for resistance, R₁, R₂and R₃ respectively, we get

So substituting these values of I₁, I₂and I₃ in eq. (i), we get
Question 4
4 / -1

The potential difference between A and B in figure is

A
1V

B
2V

C
3V

D
4V

Solution

Resistance of the upper arm CAD = 2Ω + 3Ω = 5Ω
Resistance of the lower arm CBD = 3Ω + 2Ω = 5Ω
As the resistance of both arm are equal, therefore same amount of current flows in both the arms. Current through each arm. CAD or CBD = 1A
Potential difference across C and A is V_C − V_A = (2Ω)(1A) = 2V...(i)
Potential difference across C and B is V_C − V_B = (3Ω)(1A) = 3V...(ii)
Substracting (i) from (ii), we get
V_A − V_B = 3V − 2V = 1V
Question 5
4 / -1

In the given circuit the potential at point B is zero, the potential at points A and D will be

A
V_A= 4V ; V_D = 9V

B
V_A = 3V; V_D = 4V

C
V_A = 9V; V_D = 3V

D
V_A= 4V ; V_D= 3V

Solution

V_A − V_B = 2 × 2 = 4V
∴ V_A − 0 = 4V
⇒ V_A = 4V
According to question V_B= 0
Point D is connected to positive terminal of battery of emf 3V.
Question 6
4 / -1

A current of 6A enters one corner P of an equilateral triangle PQR having 3 wires of resistances 2 Ω each and leaves by the corner R. Then the current I₁ and I₂ are

A
2 A, 4 A

B
4 A, 2 A

C
1A, 2 A

D
2A, 3 A

Solution

From Kirchhoff's first law at jucntion P
I₁+ I₂ = 6…(i)
From Kirchhoff's second law to the closed circuit PQRP,
−2I₁ − 2I₁ + 2I₂= 0
⇒ −4I₁ + 2I₂= 0
⇒ 2I₁ − I₂ = 0
Adding Eqs. (i) and (ii), we get
3I₁ = 6
⇒ I₁ = 2A
From Eq. (i),
I₂ = 6 − 2 = 4A
Question 7
4 / -1

A 7V battery with internal resistance 3Ω and 3V battery with internal resistance 10mega are connected to a 10Ω resistors as shown in figure, the current in 10Ω resistor is

A
0.27A

B
0.31A

C
0.031A

D
0.53

Solution

Using Kirchoff's law in loop AP₂P₁DA
∴ 10I₁ + 2I − 7 = 0
10I₁ + 2I = 7...(i)
Using Kirchhoff's law in loop P₂P₁CBP₂
−3 + I(I − I₁) − 10I₁= 0

I − 11I₁ = 3, I = 3 + 1₁I₁....(ii)
From (i) and (ii)
10I₁ + 2(3 + 11I₂) = 710I₁ + 6 + 22I₁ = 7
∴ 32I₁= I, I₁ = 1/32 = 0.031A
Question 8
4 / -1

In the circuit shown, the value of currents I₁, I₂ and I₃ are

A

B

C
5A, 4A, −3A

D

Solution

Applying Kirchhoff's voltage law,

In loop I,
−27 − 6I₂ − 2I₁ + 24=0
6I₂ + 2I₁= −3…(i)
In loop II,
−27 − 6I₂ + 4I, + 24 = 0
6I₂ − 4I₃ = −27…(ii)
At junction P, I₁ − I₂ − I₃ = 0…(iii)
Solving equations (i), (ii) and (iii) we get
I₁ = 3A, I₂ = −3/2A, I₃ = 9/2A.
Question 9
4 / -1

In the circuit shown, current flowing through 25V cell is

A
7.2 A

B
10 A

C
12 A

D
14.2 A

Solution

Applying KVL in loop
ABCDA, ABFEA, ABGHA and ABJIA, we get
30 − i₁ × 11 = −25… (i)
20 + i₂ × 5 = 25... (ii)
5 − i₃ × 10 = −25… (iii)
10 + i₄ × 5 = 25... (iv)
Solving equations (i), (ii), (iii) and (iv) we get
i₁ = 5A, i₂ = 1A, i₃ = 3A and i₄ = 3A
Hence, current flowing through 25V cell is 12A.
Question 10
4 / -1

A battery, an open switch and a resistor are connected in series as shown in figure.

Consider the following three statements concerning the circuit. A voltmeter will read zero if it is connected across points
(i) P and T
(ii) P and Q
(iii) Q and T
Which one of the above is/are true?

A
only (i)

B
only (iii)

C
only (i) and (iii)

D
(i), (ii) and (iii)

Solution

When the switch is not closed, a voltmeter connected across P and T will not show any potential difference.

Between Q and T also there is no potential difference because circuit is not complete.
Therefore in both the cases, the voltmeter will read zero. Between P and Q, the emf of the battery will be given.