The internal resistance of a cell of emf 2 V is 0.1 Ω. It is connected to a resistance of 3.9 Ω.
The voltage across the cell will be
(A) 0.5 V
(B) 1.5 V
(C) 1.95 V
(D) 2 V
1 Answers
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The internal resistance of a cell of emf 2 V is 0.1 Ω, it is connected to a resistance of 0.9 Ω. The voltage across the cell will be
Correct answer is (B) 1.8 V
2 Answers 3 views
A resistance of `2Omega` is connected across one gap of a metre-bridge(the length of the wire is 100 cm) and an unknown resistance, greater than `2Ome
Answer A) 3 ohm Soln-: 2/x=y/(100-x) 200-2x=xy ......1 After interchanging Y/x+20 = 2/(80-x) 80y-xy=2x+40 .......2 On solving equation 1 and 2 Y=3 ohm
2 Answers 1 views
A voltmeter is connected across a battery of emf 12 V and internal resistance of 10 Ω. If the voltmeter resistance is 230 Ω, what reading
Given: E = 12 volt, r = 10 Ω, R = 230 Ω To find: Reading shown by voltmeter (V) Formula: i. I = \(\frac{E}{R+r}\) ii. V = E – Ir Calculation: From formula (i), I = \(\frac{12}{230+10}=\frac{12}{240}=\frac{1}{20}\) A From...
2 Answers 1 views
If n identical cells, each of emf E and internal resistance r, are connected in series, write an expression for the terminal p.d.
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2 Answers 1 views
A cell of emf E and internal resistance r is connected across an external resistance R (R >> r). The p.d. across R is
Correct option is: (B) E(I – \(\frac{r}{R}\))
2 Answers 1 views
Emf of a cell is 2.2 volt. When resistance R = 5 Ω is connected in series, potential drop across the cell becomes 1.8 volt.
Correct option is: (A) 10/9 Ω
2 Answers 1 views
If a battery of emf 2 V with internal resistance one ohm is connected to an external circuit of resistance R across it
Correct option is: (D) 3 Ω
2 Answers 1 views
The emf of a cell is 12 V. When it sends a current of 1 A through an external resistance, the p.d. across the terminals reduces to 10 V.
Correct option is: (D) 2 Ω
2 Answers 1 views
The terminal voltage across a cell is more than its e.m.f., if another cell of
(A) higher e.m.f. is connected parallel to it.
2 Answers 1 views
The emf of a cell is `Ev`, and its its internal resistance is `1Omega`. A resistance of `4Omega` is joined to battery in parallel. This is connected i
Correct Answer - C `r=R[(l_(1)-l_(2))/(l_(2))]`
2 Answers 1 views