The current $I$ in the circuit shown below is :
(All diodes are ideal and identical. 
To find the current \( I \) in the circuit shown, we need to analyze the configuration of resistors and diodes. All diodes are ideal; therefore, they only allow current to flow in one direction (forward-biased) with no voltage drop across them.
Given the circuit, the diodes are oriented such that they only allow current to pass in the direction depicted by the direction of the current \( I \).
Let's analyze each branch:
- The diode in the upper branch is forward-biased, allowing current to pass through the 4Ω resistor.
- Current through the 4Ω resistor: it contributes to the total current with no obstruction.
- The diode in the middle branch is also forward-biased, allowing current through the 3Ω resistor.
- Current through the 3Ω resistor: it adds to the net current.
- The diode in the lower branch is forward-biased, allowing current through the 2Ω and 5Ω resistors in series.
- Current through the 2Ω + 5Ω resistors: contributes to the total current.
In total, each branch allows a portion of the current to contribute to the net current. Applying Kirchhoff's Voltage Law gives:
- Voltage across each resistor is the same due to diodes being ideal: \( V = 10 \, \text{V} \).
The equivalent resistance \( R_{\text{eq}} \) can be calculated by treating each branch separately:
- Upper branch: \( 4 \, \text{Ω} \)
- Middle branch: \( 3 \, \text{Ω} \)
- Lower branch: equivalent of 2Ω and 5Ω in series = 7Ω
This makes them in parallel:
\(\frac{1}{R_{\text{eq}}} = \frac{1}{4} + \frac{1}{3} + \frac{1}{7}\)
Calculating this gives:
\(R_{\text{eq}} = \frac{84}{41} \, \text{Ω} \approx 2.05 \, \text{Ω}\)
Now calculating the total current \( I \):
\(I = \frac{V}{R_{\text{eq}}} = \frac{10 \, \text{V}}{\frac{84}{41} \, \text{Ω}} = \frac{410}{84} \approx \frac{15}{2} \, \text{A}\)
Therefore, the correct answer is:
- \(\frac{15}{2} \, \text{A}\)
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Important Questions from Electricity and Magnetism
- A short bar magnet placed with its axis at $30^\circ$ with an external field of 800 Gauss, experiences a torque of 0.016 N.m. The work done in moving it from most stable to most unstable position is $\alpha \times 10^{-3} \text{ J}$. The value of $\alpha$ is ______.
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$\left[ \text{Take } \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\text{ Nm}^2\text{/C}^2 \right]$ - An inductor stores $16\text{ J}$ of magnetic field energy and dissipates $32\text{ W}$ of thermal energy due to its resistance when an a.c. current of $2\text{ A}$ (rms) and frequency $50\text{ Hz}$ flows through it. The ratio of inductive reactance to its resistance is _________. ($\pi = 3.14$)
- A simple pendulum made of mass 10 g and a metallic wire of length 10 cm is suspended vertically in a uniform magnetic field of 2 T. The magnetic field direction is perpendicular to the plane of oscillations of the pendulum. If the pendulum is released from an angle of $60^\circ$ with vertical, then maximum induced EMF between the point of suspension and point of oscillation is ___________ mV. (Take $g = 10\text{ m/s}^2$)
Using a variable frequency a.c. voltage source the maximum current measured in the given LCR circuit is 50 mA for $V = 5\sin(100t)$ The values of L and R are shown in the figure. The capacitance of the capacitor (C) used is ___________ $\mu\text{F}$.
