If a copper wire is stretched to make it 0.1% longer, the percentage increase in resistance will be:
Resistance of a stretched wire of constant volume is proportional to the square of its length: \( R \propto l^2 \). Thus, \( \frac{\Delta R}{R}\text{ (%)} = 2 \frac{\Delta l}{l}\text{ (%)} = 2 \times 0.1\% = 0.2\% \).
A galvanometer has a coil of resistance \( 100\ \Omega \) and gives full scale deflection for 20 mA current. If it is to work as a voltmeter of 10 volt range, the resistance required to be added will be:
To convert a galvanometer to a voltmeter, a high series resistance \( R \) is added: \( R = \frac{V}{I_g} - G \). Substituting values, \( R = \frac{10}{20 \times 10^{-3}} - 100 = 500 - 100 = 400\ \Omega \).
When a resistance of 2 ohm is connected across the terminals of a cell, the current is 0.5 amp. When the resistance is increased to 5 ohm, the current is 0.25 amp. The emf of the cell is:
Using \( E = I(R + r) \), we set up equations: \( E = 0.5(2 + r) \) and \( E = 0.25(5 + r) \). Equating them gives \( r = 1\ \Omega \), which yields \( E = 0.5(2 + 1) = 1.5\text{ V} \).
Which of the following is not possible by combination of four resistors each equal to \(8 \Omega \)?
With four \(8 \Omega \) resistors, the minimum possible equivalent resistance is when all four are in parallel: \(R_{min} = 8/4 = 2 \Omega\). Thus, \(1\Omega\) is impossible.
In a wire of cross-section radius \(r\), free electrons travel with drift velocity \(v\) when a current \(I\) flows through the wire. What is the current in another wire of half the radius and of the same material when the drift velocity is \(2v\)?
Since \(I = n e A v_d = n e \pi r^2 v\), the new current is \(I' = n e \pi \left(\frac{r}{2}\right)^2 (2v) = \frac{1}{2} n e \pi r^2 v = I/2\).
Arrange the following materials in increasing order of their resistivity:
Nichrome, Copper, Germanium, Silicon
Copper is a conductor (lowest resistivity), Nichrome is an alloy (intermediate), Germanium and Silicon are semiconductors with Silicon having a larger bandgap and thus higher resistivity.
Five resistors, each of resistance \(R\) are connected in series to an ideal battery. When these resistors are connected in parallel to the same battery, the current through the battery is increased \(n\) times. The value of \(n\) is
The equivalent resistance in series is \(R_s = 5R\) and in parallel is \(R_p = \frac{R}{5}\). Since current \(I = \frac{V}{R_{eq}}\), we have \(\frac{I_p}{I_s} = \frac{R_s}{R_p} = \frac{5R}{R/5} = 25\). Thus, \(n = 25\).
The reciprocal of resistance is
By definition, the reciprocal of electrical resistance is electrical conductance, \(G = \frac{1}{R}\).
Column-I gives certain physical terms associated with flow of current through a metallic conductor. Column-II gives some mathematical relations involving electrical quantities. Match Column-I and Column-II with appropriate relations.
Column-I
(A) Drift Velocity
(B) Electrical Resistivity
(C) Relaxation Period
(D) Current Density
Column-II
(P) \(\frac{m}{n e^2 rho}\)
(Q) \(n e v_d\)
(R) \(\frac{e E}{m} \tau\)
(S) \(\frac{E}{J}\)
Choose the correct match from the given options:
Using standard formulas of current electricity: drift velocity \(v_d = \frac{eE}{m}\tau\) gives (A)-(R); resistivity \(\rho = \frac{E}{J}\) gives (B)-(S); relaxation time \(\tau = \frac{m}{ne^2\rho}\) gives (C)-(P); current density \(J = ne v_d\) gives (D)-(Q).
The effective resistance of a parallel connection that consists of four wires of equal length, equal area of cross section and same material is 0.25 \(\Omega\). What will be the effective resistance if they are connected in series?
For four identical wires of resistance \(R\) connected in parallel, \(R_p = \frac{R}{4} = 0.25, \Omega ⇒ R = 1, \Omega\). In series, \(R_s = 4R = 4 \times 1, \Omega = 4, \Omega\).