Current Electricity

The drift velocity $v_d$ is related to the potential difference V by the equation:$v_d=\frac{eV\zeta }{ml}$.

$v_d\alpha V$
Thus, when the potential difference is halved, the drift velocity will also be halved

\(\vec{E}\) and \(\vec{j}\) both are along "-ve" X axis direction.

\(\vec{E}\) and \(\vec{j}\) both are along "-ve" X axis direction.

New Length \(L' = \frac{L}{2}\), New Radius \(r' = r\).

(This is correct).

The drift velocity $v_d$ is related to the potential difference V by the equation:$v_d=\frac{eV\zeta }{ml}$.

$v_d\alpha V$
Thus, when the potential difference is halved, the drift velocity will also be halved

\(R=\frac{\rho L}{A}\)

For wire P , the resistance is given as \(R=\frac{\rho L}{A}\).
Since wire Q has twice the diameter of wire P , its cross-sectional area \(A_Q\) is four times that of P (because area \(A \propto d^2\) ). Thus,

\(A_Q=4 A\)

Wire Q also has half the length of wire P , so \(L_Q=\frac{L}{2}\).
The resistance of wire \(\mathrm{Q}, R_Q\), is:

\(R_Q=\frac{\rho L_Q}{A_Q}=\frac{\rho\left(\frac{L}{2}\right)}{4 A}=\frac{\rho L}{8 A}=\frac{R}{8}\)

The equivalent EMF of two non-ideal batteries connected in parallel is given by:

$E_{cq}=\frac{E_1{r}_2+E_2{r}_1}{r_1+r_2}$

where $E_1,E_2$ are the EMFs of the two batteries and $r_1,r_2$ are their internal resistances.

When a galvanometer is converted to an ammeter, a shunt resistance $S$ is connected in parallel with the galvanometer. The shunt allows most of the current to bypass the galvanometer.

1. Full-scale current through galvanometer ($I_g$):

   $$I_g = 20 \, \text{mA} = 0.02 \, \text{A}$$

2. Maximum current through the ammeter ($I$):

   $$I = 3 \, \text{A}$$I=3A

3. Current through the shunt ($I_s$​):

   $$I_s = I - I_g = 3 - 0.02 = 2.98 \, \text{A}$$

4. Relationship between shunt resistance and galvanometer resistance:
   Since $S$ and $G$ are in parallel, the voltage across both is the same:

   $$I_s S = I_g G$$

   Substituting the known values:

   $$2.98 S = 0.02 \times 30$$$$2.98 S = 0.6$$

5. Solve for $S$:

   $$S = \frac{0.6}{2.98} \approx 0.2 \, \Omega$$

Thus, the value of the shunt resistance is:

$$\boxed{0.2 \, \Omega}$$​

\(R=\frac{\rho L}{A}\)

For wire P , the resistance is given as \(R=\frac{\rho L}{A}\).
Since wire Q has twice the diameter of wire P , its cross-sectional area \(A_Q\) is four times that of P (because area \(A \propto d^2\) ). Thus,

\(A_Q=4 A\)

Wire Q also has half the length of wire P , so \(L_Q=\frac{L}{2}\).
The resistance of wire \(\mathrm{Q}, R_Q\), is:

\(R_Q=\frac{\rho L_Q}{A_Q}=\frac{\rho\left(\frac{L}{2}\right)}{4 A}=\frac{\rho L}{8 A}=\frac{R}{8}\)

Sol.

• Statement I: This is false. When non-ideal batteries are connected in parallel, the resultant EMF is a weighted average of the EMFs of the individual batteries depending on their internal resistances. It does not necessarily become less than either battery's EMF.

• Statement II: This is true. When internal resistances are connected in parallel, the resultant resistance is always smaller than the smallest individual resistance, as per the formula for parallel resistance:

  $$\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2}.$$

Thus, the correct answer is $\text{(b)}$.

• Total resistance of the wire: $R_{\text{total}} = 9 \, \Omega$
• Resistance of each side: $R_{\text{side}} = \frac{R_{\text{total}}}{3} = 3 \, \Omega$
• Between any two vertices, there are two paths:
  1. Direct resistance $R_1 = 3 \, \Omega$.
  2. The other two sides ($R_2$) add up: $R_2 = 6 \, \Omega$
• These are in parallel: $$R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2} = \frac{(3)(6)}{3 + 6} = 2 \, \Omega.$$

Thus, the equivalent resistance is $2 \, \Omega$.

The drift velocity $v_d$ is related to the potential difference V by the equation:$v_d=\frac{eV\zeta }{ml}$.

$v_d\alpha V$
Thus, when the potential difference is halved, the drift velocity will also be halved

New Length \(L' = \frac{L}{2}\), New Radius \(r' = r\).

(This is correct).

\(\vec{E}\) and \(\vec{j}\) both are along "-ve" X axis direction.

• Total resistance of the wire: $R_{\text{total}} = 9 \, \Omega$
• Resistance of each side: $R_{\text{side}} = \frac{R_{\text{total}}}{3} = 3 \, \Omega$
• Between any two vertices, there are two paths:
  1. Direct resistance $R_1 = 3 \, \Omega$.
  2. The other two sides ($R_2$) add up: $R_2 = 6 \, \Omega$
• These are in parallel: $$R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2} = \frac{(3)(6)}{3 + 6} = 2 \, \Omega.$$

Thus, the equivalent resistance is $2 \, \Omega$.

Let the unknown resistance in the right gap be \(R\). By the meter bridge principle:

\(\frac{2}{40}=\frac{R}{100-40}\Longrightarrow \frac{2}{40}=\frac{R}{60}\Longrightarrow R=3\Omega\)

When \(R\) is shunted (in parallel) with \(2\Omega\), the new right gap resistance is:

\(R'=\frac{3\times2}{3+2}=\frac{6}{5}=1.2\Omega\)

Let the new balancing length be \(l'\):

\(\frac{2}{l'}=\frac{1.2}{100-l'}\Longrightarrow 200-2l'=1.2l'\Longrightarrow 3.2l'=200\Longrightarrow l'=62.5\ \text{cm}\)

The change in balancing length is \(62.5\ \text{cm}-40\ \text{cm}=22.5\ \text{cm}\).

New Length \(L' = \frac{L}{2}\), New Radius \(r' = r\).

(This is correct).

New Length \(L' = \frac{L}{2}\), New Radius \(r' = r\).

(This is correct).

The drift velocity $v_d$ is related to the potential difference V by the equation:$v_d=\frac{eV\zeta }{ml}$.

$v_d\alpha V$
Thus, when the potential difference is halved, the drift velocity will also be halved

\(\vec{E}\) and \(\vec{j}\) both are along "-ve" X axis direction.

$$\begin{gathered} S=\frac{i_{g}G}{i-i_g} \\ S=\frac{20\times {10}^{-3}\times 30}{2.98} \\ S=\frac{20\times 30}{2980} \\ S=\frac{30}{149}\Omega \end{gathered}$$

\(R=\frac{\rho L}{A}\)

For wire P , the resistance is given as \(R=\frac{\rho L}{A}\).
Since wire Q has twice the diameter of wire P , its cross-sectional area \(A_Q\) is four times that of P (because area \(A \propto d^2\) ). Thus,

\(A_Q=4 A\)

Wire Q also has half the length of wire P , so \(L_Q=\frac{L}{2}\).
The resistance of wire \(\mathrm{Q}, R_Q\), is:

\(R_Q=\frac{\rho L_Q}{A_Q}=\frac{\rho\left(\frac{L}{2}\right)}{4 A}=\frac{\rho L}{8 A}=\frac{R}{8}\)