In the figure shown below, a resistance of 150.4 $ \Omega $ is connected in series to an ammeter A of resistance 240 $ \Omega $. A shunt resistance of 10 $ \Omega $ is connected in parallel with the ammeter. The reading of the ammeter is ______ mA.
In the figure shown below, a resistance of 150.4 $ \Omega $ is connected in series to an ammeter A of resistance 240 $ \Omega $. A shunt resistance of 10 $ \Omega $ is connected in parallel with the ammeter. The reading of the ammeter is ______ mA.
Show Hint
Correct Answer: 5
Approach Solution - 1
This problem involves a circuit with a voltage source, a series resistor, and an ammeter shunted by another resistor. We need to find the reading of the ammeter, which is the current flowing through it.
Concept Used:
The solution requires the application of several fundamental circuit laws:
- Equivalent Resistance in Parallel: For two resistors \( R_1 \) and \( R_2 \) in parallel, the equivalent resistance \( R_p \) is given by: \[ R_p = \frac{R_1 R_2}{R_1 + R_2} \]
- Equivalent Resistance in Series: For resistors in series, the total resistance is the sum of individual resistances.
- Ohm's Law: The total current \( I \) flowing in a circuit is given by \( I = \frac{V}{R_{eq}} \), where \( V \) is the total voltage and \( R_{eq} \) is the total equivalent resistance of the circuit.
- Current Divider Rule: When a total current \( I_{total} \) enters a parallel combination of two resistors \( R_1 \) and \( R_2 \), the current \( I_1 \) through resistor \( R_1 \) is: \[ I_1 = I_{total} \times \frac{R_2}{R_1 + R_2} \]
Step-by-Step Solution:
Step 1: Calculate the equivalent resistance of the parallel combination of the ammeter and the shunt resistor.
The ammeter has a resistance \( R_A = 240 \, \Omega \), and the shunt resistor has a resistance \( R_{sh} = 10 \, \Omega \). These two are connected in parallel. \[ R_p = \frac{R_A \times R_{sh}}{R_A + R_{sh}} \] \[ R_p = \frac{240 \times 10}{240 + 10} = \frac{2400}{250} = 9.6 \, \Omega \]
Step 2: Calculate the total equivalent resistance of the entire circuit.
The series resistor \( R_s = 150.4 \, \Omega \) is connected in series with the parallel combination \( R_p \). \[ R_{eq} = R_s + R_p \] \[ R_{eq} = 150.4 + 9.6 = 160 \, \Omega \]
Step 3: Calculate the total current flowing from the voltage source.
The voltage of the source is \( V = 20 \, \text{V} \). Using Ohm's Law: \[ I_{total} = \frac{V}{R_{eq}} = \frac{20}{160} = \frac{1}{8} \, \text{A} \] \[ I_{total} = 0.125 \, \text{A} \]
Step 4: Calculate the current flowing through the ammeter using the current divider rule.
The total current \( I_{total} \) splits between the ammeter (\( R_A \)) and the shunt resistor (\( R_{sh} \)). The reading of the ammeter is the current flowing through it, \( I_A \). \[ I_A = I_{total} \times \frac{R_{sh}}{R_A + R_{sh}} \] \[ I_A = 0.125 \times \frac{10}{240 + 10} \] \[ I_A = 0.125 \times \frac{10}{250} = 0.125 \times \frac{1}{25} \] \[ I_A = \frac{0.125}{25} = 0.005 \, \text{A} \]
Step 5: Convert the ammeter reading to milliamperes (mA).
To convert from amperes to milliamperes, we multiply by 1000. \[ I_A = 0.005 \, \text{A} \times 1000 \, \frac{\text{mA}}{\text{A}} = 5 \, \text{mA} \]
The reading of the ammeter is 5 mA.
Approach Solution -2
$R_{eq} = R_1 + R_2$
$R_{eq} = 150.4 + \frac{240 \times 10}{250}$
$R_{eq} = 150.4 + 9.6 = 160 \Omega$
$I_1 = \frac{IR_2}{240}$
$I_1 = \frac{I \times 9.6}{240}$
$I = \frac{20}{160}$
$I_1 = \frac{20}{160} \times \frac{9.6}{240} = \frac{1}{200} = 5 \times 10^{-3} A = 5 mA$
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
Top JEE Main Current electricity Questions
- In the given circuit, find electric current drawn from battery :
- If the current through an incandescent lamp decreases by 20%, how much change will be there in its illumination?
- A potential divider circuit is shown in figure. The output voltage V0 is
- Two cells are connected in opposition as shown. Cell \( E_1 \) is of 8 V emf and 2 \( \Omega \) internal resistance; the cell \( E_2 \) is of 2 V emf and 4 \( \Omega \) internal resistance. The terminal potential difference of cell \( E_2 \) is:
- Wheatstone bridge principle is used to measure the specific resistance \( (S_1) \) of a given wire, having length \( L \), radius \( r \). If \( X \) is the resistance of the wire, then specific resistance is: $$ S_1 = X\left( \frac{\pi r^2}{L} \right). $$ If the length of the wire gets doubled, then the value of specific resistance will be:
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.