Current Electricity
- The rate of increase of thermo–e.m.f. with temperature at the neutral temperature of a thermocouple
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We have,
e = at + bt²
⇒ de/dt = a + 2bt
At neutral temperature,
t = -(a/2b)
∴ de/dt = 0Correct Option: B
We have,
e = at + bt²
⇒ de/dt = a + 2bt
At neutral temperature,
t = -(a/2b)
∴ de/dt = 0
- A current of 2A flows through a 2Ω resistor when connected across a battery. The same battery supplies a current of 0.5 A when connected across a 9Ω resistor. The internal resistance of the battery is
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Let the internal resistance of the battery be r. Then the current flowing through the circuit is given by
i = E R + r
In first case,2 = E ...(i) 2 + r
In second case,0.5 = E ...(i) 9 + r
From (1) & (2), 4 + 2r = 4.5 + 0.5 r
⇒1.5 r = 0.5 ⇒ r = 1/3 Ω.Correct Option: B
Let the internal resistance of the battery be r. Then the current flowing through the circuit is given by
i = E R + r
In first case,2 = E ...(i) 2 + r
In second case,0.5 = E ...(i) 9 + r
From (1) & (2), 4 + 2r = 4.5 + 0.5 r
⇒1.5 r = 0.5 ⇒ r = 1/3 Ω.
- Cell having an emf ε and internal resistance r is connected across a variable external resistance R. As the resistance R is increased, the plot of potential difference V across R is given by :
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The current through the resistance R
I = 
R ε R + r
The potential difference across RV = IR = R ε R + r 
when R = 0, V = 0,
R = ∞, V = ε Thus V increases as R increases upto certain limit, but it does not increase further.
Correct Option: C
The current through the resistance R
I = R ε R + r
The potential difference across RV = IR = R ε R + r
when R = 0, V = 0,
R = ∞, V = ε Thus V increases as R increases upto certain limit, but it does not increase further.
- The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10 Ω is
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Given : emf ε = 2.1 V
I = 0.2 A, R = 10Ω
Internal resistance r = ?
From formula.
ε – Ir = V = IR
2.1 – 0.2r = 0.2 × 10
2.1 – 0.2 r = 2 or 0.2 r = 0.1⇒ r = 0.1 = 0.5 Ω 0.2 
Correct Option: A
Given : emf ε = 2.1 V
I = 0.2 A, R = 10Ω
Internal resistance r = ?
From formula.
ε – Ir = V = IR
2.1 – 0.2r = 0.2 × 10
2.1 – 0.2 r = 2 or 0.2 r = 0.1⇒ r = 0.1 = 0.5 Ω 0.2
- Two batteries of emf 4 V and 8V with internal resistance 1 Ω and 2 Ω are connected in a circuit with a resistance of 9 Ω as shown in figure. The current and potential difference between the points P and Q are
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I = 8 - 4 = 4 = 1 A 1 + 2 + 9 12 3 VP - VQ = 4 - 1 × 3 = 3 volt 3
Correct Option: A
I = 8 - 4 = 4 = 1 A 1 + 2 + 9 12 3 VP - VQ = 4 - 1 × 3 = 3 volt 3
