The potential difference across a cell is 1.8 V when a current of 0.5 A is drawn from it. The p.d. falls to 1.6 V when a current of 1.0 A is drawn. Find the emf and the internal resistance of the cell.
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“The current through an inductor cannot change instantaneously. However, the potential difference across an inductor can change abruptly”. Do you agre
Correct Answer - Statement is true
2 Answers 2 views
The terminals of a cell are cnnected to resistance R and the falll of potential across R is balaced against the fall of potential on a potentiometer w
Correct Answer - 6.67 `Omega`
2 Answers 1 views
Two, capacitors `A` and `B` are connected in series across a `100 V` supply and it is observed that the potential difference across them are `60 V` an
Correct Answer - `0.16muF, 0.24 muF`
2 Answers 1 views
With a cell of emf 1.5 V, the balance point is obtained at a distance of 60 cm from one end of the potential wire. Calculate the potential difference
Correct Answer - 2.5 V
2 Answers 1 views
A standard cell of emf 1.08 V is balanced by the potential difference across 91 cm of a meter long wire supplied by a cell of emf 2 V through a series
Correct Answer - `0.03Omegacm^(-1)`
2 Answers 1 views
vii) A cell of emf \( 1.1 V \) and internal resistance \( r \) is connected across an external resistor of resistance \( R=10 \). The potential difference across the resistor is a) \( 0.1 V \) b) \( 0.9 V \) c) \( 1.0 V \) d) \( 1.1 V \)
Given E = 1.1 V r = r R = 10 r Potential difference across the resistor = IR = (E/R+r) R = (1.1/10r+r) 10 r \(=\frac{1.1\times10\,r}{11\,r\times10}\) = 11/11 = 1 volt option (c) is correct.
2 Answers 2 views
A well collimated parallel pencil of cathode rays falls through a potential difference `3kV` & enters the spacing between two parallel metallic plates
Correct Answer - `[0.05cm]`
2 Answers 4 views
The potential difference of a cell in an open circuit is 6 V, which falls to 4 V when a current of 2 A is drawn from the cell. Calculate the emf and t
Correct Answer - `6 V, 1 Omega`
2 Answers 1 views
When a resistance of `2 Omega` is placed across the terminals of a battery, the current is 0.5 A. When the resistance across the terminals of the batt
Correct Answer - `1.5 V, 1 Omega`
2 Answers 1 views
A potential difference of V = 100 ± 2 volt, when applied across a resistance R gives a current of 10 ± 0.5 ampere.
Here. V = 100 ± 2 volt and I = 10 ± 0.5 ampere Expressing limits of error as percentage error, We have, V = 100 volt ± \(\frac{2}{100}\) × 100% = 10 volt ± 2% and...
2 Answers 1 views