Lcr Circuit

To find the current flowing through the 20 Ω resistance, we must analyze the circuit in terms of the voltage across the external resistance and the total resistance in the circuit.

First, consider the two batteries in parallel:

• Battery 1: Voltage, ; internal resistance, .
• Battery 2: Voltage, ; internal resistance, .

The equivalent voltage and equivalent internal resistance for parallel batteries are given by:

Substitute the given values:

For the equivalent internal resistance:

Now, the total resistance, , in the circuit is the sum of and the external resistance, 20 :

The current through the 20 resistance can then be found using Ohm's law:

Therefore, the current flowing through the 20 Ω resistance is approximately 0.53 A.

To determine how many electrons flow past a given cross-section of a copper wire in 10.0 minutes, given the current is 80.0 mA, we need to use the formula for current, which relates charge and time:

where:

• is the current (Amperes)
• is the total charge (Coulombs)
• is the time (seconds)

First, convert the current from milliamperes to amperes:

Next, convert time from minutes to seconds:

Rearrange the formula to solve for :

Substitute the values into the equation:

Next, calculate the number of electrons using the charge of a single electron, :

Therefore, the correct answer is electrons.

To determine the mutual conductance ( ) of a triode valve, we need to use the relationship between the amplification factor ( ), anode resistance ( ), and mutual conductance. The formula is given by:

Here, we have:

• The amplification factor ( ) is 20.
• The anode resistance ( ) is 20,000 ohms.

Substituting these values into the formula gives:

Simplifying the fraction:

Thus, the mutual conductance is .

To solve the problem of determining the maximum and minimum possible intensities when two coherent monochromatic light beams are superimposed, we must understand the principle of superposition and interference.

When two light beams interfere, the resultant intensity depends on the phase difference between them. The resultant intensity is given by:

I_r =

where and are the intensities of the individual beams, and is the phase difference between the beams. Let's apply this for the given problem:

Given: and

1. **Maximum Intensity**: This occurs when the beams interfere constructively, i.e., . The cosine term becomes 1.

=
=

2. **Minimum Intensity**: This occurs when the beams interfere destructively, i.e., . The cosine term becomes -1.

=
=

Thus, the maximum and minimum possible intensities in the resulting beam are and respectively.