Electric Circuits Test 1

Given:

• Voltage source: 20V

• Resistances: R1 = 2Ω and R2 = 3Ω

• Configuration: The resistors are connected in series with a voltage source of 20V.

1. Determine the Total Resistance: Since the resistors are in series: R_total = R1 + R2 = 2Ω + 3Ω = 5Ω

2. Determine the Total Current: Using Ohm's law, we calculate the current in the circuit: I = V_total / R_total = 20V / 5Ω = 4A

3. Determine the Voltage Drop across Each Resistor: The voltage drop across each resistor is given by Ohm's law:

   • Voltage drop across R1:
     V1 = I × R1 = 4A × 2Ω = 8V

   • Voltage drop across R2:
     V2 = I × R2 = 4A × 3Ω = 12V

4. Direction of the Voltage Drop:

   • The current flows from the positive terminal of the 20V battery, passing through R1 and then through R2.

   • Since resistors are passive elements, the voltage drop across each resistor reduces the potential as current flows. The voltage drop across R2 (from the perspective of the battery) will be negative because the current flows from the positive terminal to the negative terminal of the resistor.

This results in: V2 = -12V

Diode is non-linear elements is removed.

After shorting the voltage source the redrawn circuit is

Thevenin resistance

= 11/3 ohm

Given that

L = 50 × 10^-3 H

C = 5 × 10^-6 F

Resonant frequency, \({\omega _0} = \frac{1}{{\sqrt {LC} }}\)

\({\omega _0} = \frac{1}{{\sqrt {50 \times {{10}^{ - 3}} \times 5 \times {{10}^{ - 6}}} }} = 2000rad/sec\;\;\)

Given that, current = 10 mA

At resonance, \({I_0} = \frac{V}{R}\)

\(\Rightarrow 10 \times {10^{ - 3}} = \frac{{50}}{R}\)

⇒ R = 5 KΩ

Quality factor, \(Q = \frac{{{\omega _0}L}}{R} = \frac{{2000 \times 50 \times {{10}^{ - 3}}}}{{5 \times {{10}^3}}} = 0.02\)

• Based on kirchoffs current law, we know that incoming currents at a node is equals to outgoing currents.
• So, based on this 4 and 3 will get canceled with 5 and 2.
• So, finally i = 1.

In order for 600 C charge to be delivered to the 100 V source, the current must be

Going from 10 V to 0 V:

Minus sign indicates that the polarity of battery should be reversed.

The relationship between electric potential difference (V), work done (W), and charge (Q) is given by the formula:

V = W / Q

Rearranging for Q, we get:

Q = W / V

Substituting the given values:

• W = 660 J
• V = 110 V

Calculating Q:

Q = 660 J / 110 V = 6 C

Thus, the correct answer is B.

The current i will be distributed in the cube branches symmetrically

If we go from +ve side of 1 kΩ through 7 V, 6 V and 5 V

We get, v₁ = 7 + 6 - 5 = 8 V

Let's assume the current source as the ideal source:

Apply KVL in the loop, we get

v - 4 x 2.5 - 10 = 0

⇒ v = 20 V

• Cosɸ = R/Z; where R= 10 ohms and Z = 20 ohms  
• Cosɸ = 1/2 means phase angle is 60 degree.

We know that,
I=Q / t or Q = I x t = 10 x 1 hours
{since unit of current is Cs^-1, therefore time should be in seconds}
∴ Q = 10 x 60 x 60
= 36000 C
= 3.6 x 10⁴ C

This 0.2 V increases linearly from 0 to 0.2 V. Then current is zero. So capacitor hold this voltage.

we know that,

Cut – off frequency \({f_c} = \frac{1}{{2\pi RC}}\)

\(\begin{array}{l} \Rightarrow R = \frac{1}{{2\pi {f_c}C}}\\ \Rightarrow R = \frac{1}{{2\pi {f_c}C}}\\ \Rightarrow R = \frac{1}{{2\pi \times 1500 \times 500 \times {{10}^{ - 9}}}} \end{array}\)

⇒ R = 212.2 Ω

Given 3-phase system is balanced, let the instantaneous voltage is

V_R (t) = V_m cosωt

V_Y (t) = V_m cos(ωt - 120°)

V_B (t) = V_m cos(ωt + 120°)

V_m cosωt = 60 V

V_m cos(ωt - 120°) = -120 V

⇒ V_m [cosωt cos120 + sinωt sin120] = -120

\(\begin{array}{l} \Rightarrow {V_m}\left[ {{\rm{cos\omega t}}\left( { - \frac{1}{2}} \right) + sin\omega t\left( {\frac{{\sqrt 3 }}{2}} \right)} \right] = - 120\\ \Rightarrow 60\left( { - \frac{1}{2}} \right) + \frac{{\sqrt 3 }}{2}{V_m}sin\omega t = - 120\\ \Rightarrow 60\left( { - \frac{1}{2}} \right) + \frac{{\sqrt 3 }}{2}{V_m}sin\omega t = - 120 \Rightarrow 60\left( { - \frac{1}{2}} \right) + \frac{{\sqrt 3 }}{2}{V_m}sin\omega t = - 120\end{array}\)  

V_m cosωt = 60 V

⇒ tanωt = - √3 ⇒ ωt = -60°

\(\begin{array}{l}V_m^2{\sin ^2}\omega t + V_m^2{\cos ^2}\omega t = {\left( {60\sqrt 3 } \right)^2} + {\left( {60} \right)^2} \Rightarrow {V_m} = 120\;V\\{V_B} = {V_m}\cos \left( {\omega t + 120} \right) = 120\cos \left( { - 60 + 120} \right) = 120 \times \frac{1}{2} = 60\;V\end{array}\)