Q.Two batteries of emf $\varepsilon_1$ and $\varepsilon_2$ (with $\varepsilon_2 > \varepsilon_1$) and internal resistances $r_1$ and $r_2$ respectively are connected in parallel between two common terminals $A$ and $B$: one branch is the first battery $\varepsilon_1$ in series with $r_1$, and the other branch is the second battery $\varepsilon_2$ in series with $r_2$, both branches joining the same two terminals $A$ and $B$ with their positive terminals on the same side. Which of the following statements about the equivalent emf $\varepsilon_{eq}$ of this combination is correct?

Given

• Voltmeter reading $V = 100~\text{V}$
• Voltmeter resistance $R_V = 200~\Omega$

The current flowing through the voltmeter branch is

This same current flows through $R_2$ (it is in series with the voltmeter in the shown segment). Working the loop with the supply and series elements shown in the figure, the drop across $R_2$ is $V_2 = i\,R_2$, and solving the circuit gives

A galvanometer is a sensitive instrument designed to detect very small currents. It gives a full-scale deflection for a specific, usually small, current. To convert it into an ammeter capable of measuring larger currents, we need to ensure two things:

1. Protection: The galvanometer itself should not be damaged by the large current. Only the small current required for its full-scale deflection ($I_g$) should pass through it.
2. Range Extension: The instrument as a whole should be able to measure the desired maximum current ($I$).
3. Low Resistance: An ideal ammeter has zero resistance. A practical ammeter must have a very low resistance so that when connected in series in a circuit, it does not significantly alter the circuit's total resistance and thus the current it is trying to measure.

To achieve this, a very low resistance, called a shunt resistance ($R_s$), is connected in parallel with the galvanometer.

Why parallel connection and low resistance?

When the shunt resistance is connected in parallel:

• The majority of the large current ($I - I_g$) bypasses the galvanometer and flows through the shunt.
• Only the small current ($I_g$) required for full-scale deflection passes through the galvanometer.
• The equivalent resistance of the parallel combination of the galvanometer and the shunt is very low, which is essential for an ammeter.

This setup allows the galvanometer to indicate the total current flowing through the parallel combination, effectively extending its range without damaging it.

Here's how to calculate the required shunt resistance:

1. Identify the given parameters:

   • Resistance of the galvanometer, $R_g = 99.9~\Omega$.
   • Current for full-scale deflection of the galvanometer, $I_g = 5~\mathrm{mA} = 5 \times 10^{-3}~\mathrm{A}$.
   • Desired range of the ammeter (maximum current it should measure), $I = 5~\mathrm{A}$.

2. Understand the circuit configuration:

   When the galvanometer is converted into an ammeter, the shunt resistance $R_s$ is connected in parallel with the galvanometer $R_g$. The total current $I$ to be measured enters this parallel combination.

3. Apply Kirchhoff's Current Law (KCL):

   The total current $I$ splits into two parts: $I_g$ flowing through the galvanometer and $I_s$ flowing through the shunt.

Therefore, the current flowing through the shunt is:

4. Apply Kirchhoff's Voltage Law (KVL) for parallel branches: In a parallel connection, the voltage across each branch is the same. So, the voltage across the galvanometer is equal to the voltage across the shunt resistance.

5. Express voltages using Ohm's Law ($V=IR$):

6. Substitute $I_s$ from Step 3 into the equation from Step 5:

7. Solve for the shunt resistance $R_s$:

> [!FORMULA]
> The shunt resistance $R_s$ required to convert a galvanometer into an ammeter is given by:
> $$R_s = \frac{I_g R_g}{I - I_g}$$
> where $I_g$ is the full-scale deflection current of the galvanometer, $R_g$ is the galvanometer resistance, and $I$ is the desired maximum current range of the ammeter.

8. Substitute the numerical values and calculate $R_s$:

Ensure all units are consistent (Amperes for current, Ohms for resistance).

> [!WARNING]
> A common mistake is to forget to convert the galvanometer current from milliamperes (mA) to amperes (A) before calculation. Always ensure all units are consistent.

Comparing this result with the given options:

(A) $0.01~\Omega$

(B) $10~\Omega$

(C) $1~\Omega$

(D) $0.1~\Omega$

The calculated shunt resistance is $0.1~\Omega$, which matches option (D).

Concept and Intuition

When we talk about a "null point" in electrostatics, it typically refers to a point where the net electric field is zero. This is a crucial distinction from a point where the electric potential is zero.

For two point charges, the location of the null point for the electric field depends on their signs and magnitudes:

• If the charges have the same sign, the null point for the electric field will be between them.
• If the charges have opposite signs, as in this problem, the null point for the electric field will be outside the region between them, on the line connecting them, and closer to the charge with the smaller magnitude. This is because the electric fields due to the two charges must point in opposite directions and have equal magnitudes to cancel out.

Once we find this null point for the electric field, we then need to calculate the electric potential at that specific location. Electric potential is a scalar quantity, meaning it has magnitude but no direction. The total electric potential at a point due to multiple charges is simply the algebraic sum of the potentials due to each individual charge.

Step-by-step Derivation

1. Identify the given charges and separation:

   We have two charges:

   • $q_1 = -2.5\,\mu\text{C} = -2.5 \times 10^{-6}\,\text{C}$
   • $q_2 = 10\,\mu\text{C} = 10 \times 10^{-6}\,\text{C}$ The distance separating them is $d = 0.5\,\text{m}$. Coulomb's constant $k = 9 \times 10^9\,\text{N m}^2/\text{C}^2$.

2. Locate the null point for the electric field:

   Since the charges are of opposite signs ($q_1$ is negative, $q_2$ is positive), the null point for the electric field will be outside the line segment connecting them. Also, since $|q_1| = 2.5\,\mu\text{C}$ is smaller than $|q_2| = 10\,\mu\text{C}$, the null point will be closer to $q_1$.

   Let's place $q_1$ at the origin ($x=0$) and $q_2$ at $x=d = 0.5\,\text{m}$. The null point P must be to the left of $q_1$ (i.e., at $x < 0$).

   Let the distance of the null point P from $q_1$ be $x$. So, the position of P is $-x$.

   The distance from $q_1$ to P is $r_1 = x$.

   The distance from $q_2$ to P is $r_2 = d+x = 0.5+x$.

   At the null point P, the magnitude of the electric field due to $q_1$ ($E_1$) must be equal to the magnitude of the electric field due to $q_2$ ($E_2$).

   $E_1 = \frac{k|q_1|}{r_1^2}$ and $E_2 = \frac{k|q_2|}{r_2^2}$

   Setting $E_1 = E_2$:

Divide both sides by $2.5$:

Taking the square root of both sides:

We consider the positive root because $x$ represents a distance, and the null point is outside the charges, so $x$ must be positive.

So, the null point is located $0.5\,\text{m}$ to the left of $q_1$.
The distances from the charges to the null point are:
*   $r_1 = 0.5\,\text{m}$ (from $q_1$)
*   $r_2 = 0.5 + 0.5 = 1.0\,\text{m}$ (from $q_2$)

3. Calculate the electric potential at the null point:

The total electric potential $V$ at the null point is the algebraic sum of the potentials due to $q_1$ and $q_2$:

Substitute the values:

Factor out $10^{-6}$:

The calculated potential is $4.5 \times 10^4\,\text{V}$. Comparing this with the given options, it matches option (C).

When a DC voltage source is connected to a series RL circuit (a resistor and an inductor in series), the current does not instantly reach its maximum value. This is because an inductor opposes any change in the current flowing through it.

Initially, at the moment the switch is closed ($t=0$), the inductor acts like an open circuit, preventing any current from flowing. As time progresses, the current starts to build up. The inductor generates a back EMF that opposes this increase in current. Over time, this opposition diminishes, and the current gradually increases, approaching a steady-state value.

The steady-state current is reached when the inductor effectively acts as a short circuit (its resistance for DC is zero). At this point, the entire voltage drop occurs across the resistor, and the current is simply determined by Ohm's law. The rate at which the current rises is characterized by the time constant $\tau = L/R$, where $L$ is the inductance and $R$ is the resistance. A larger time constant means the current takes longer to reach its steady-state value.

The mathematical expression for the current $I(t)$ at any time $t$ in an RL circuit connected to a DC voltage $V$ is given by:

We will use this formula to find the inductance $L$.

1. Identify the given parameters:

   We are given the following values:

   • Resistance of the coil, $R = 50 \, \Omega$
   • Voltage of the battery, $V = 5.0 \, \text{V}$
   • Current in the coil at time $t = 0.1 \, \text{s}$, $I(t) = 50 \, \text{mA}$. It's crucial to convert the current to Amperes for consistency with other SI units: $50 \, \text{mA} = 50 \times 10^{-3} \, \text{A} = 0.05 \, \text{A}$.

2. Calculate the steady-state current ($I_0$):

   The steady-state current is the maximum current that flows through the circuit after a long time, when the inductor behaves like a short circuit. It can be calculated using Ohm's law:

   $I_0 = \frac{V}{R} = \frac{5.0 \, \text{V}}{50 \, \Omega} = 0.1 \, \text{A}$

3. Substitute values into the current growth formula:

   Now, we use the formula for current growth: $I(t) = I_0 \left(1 - e^{-t/\tau}\right)$.

   We know $I(t) = 0.05 \, \text{A}$, $I_0 = 0.1 \, \text{A}$, and $t = 0.1 \, \text{s}$. The time constant $\tau = L/R = L/50$.

   Substituting these values:

   $0.05 = 0.1 \left(1 - e^{-0.1/(L/50)}\right)$

4. Solve for the inductance ($L$):

   First, divide both sides by $0.1$:

   $\frac{0.05}{0.1} = 1 - e^{-0.1 \times \frac{50}{L}}$

   $0.5 = 1 - e^{-5/L}$

   Rearrange the equation to isolate the exponential term:

   $e^{-5/L} = 1 - 0.5$

   $e^{-5/L} = 0.5$

   To solve for $L$, take the natural logarithm ($\ln$) of both sides:

   $\ln\left(e^{-5/L}\right) = \ln(0.5)$

   $-5/L = \ln(0.5)$

   Recall that $\ln(0.5) = \ln(1/2) = \ln(1) - \ln(2) = 0 - \ln(2) = -\ln(2)$.

   So, the equation becomes:

   $-5/L = -\ln(2)$

   $5/L = \ln(2)$

   Finally, solve for $L$:

   $L = \frac{5}{\ln(2)}$