Engineering College

if the resistive current is 2 a and the inductive current is 2 a in a parallel rl circuit, total current is ________

Answers

Answer 1

If the resistive current is 2A and the inductive current is 2A, the total current  in the parallel RL circuit is 2.83A.

Since it is a parallel circuit, the voltage across the resistor and inductor are the same. The resistive current and inductive current can be combined to find the total current using the phasor diagram. Therefore, the total current in the parallel RL circuit is equal to the phasor sum of the resistive current and the inductive current.

The phasor diagram is a tool used to represent the resistive and inductive components of the circuit. In a phasor diagram, the resistive current and the inductive current are plotted along the X-axis and Y-axis, respectively. The total current can be calculated by adding the resistive current and inductive current in a vector manner.

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Answer 2

The total current in the parallel RL circuit is 4 A.

In a parallel RL circuit, the total current is calculated as follows:

Total Current = I1 + I2

Where I1 is the current flowing through the resistor (resistive current) and I2 is the current flowing through the inductor (inductive current).

According to the problem statement, the resistive current is 2 A and the inductive current is also 2 A.

Therefore, the total current is:Total Current = I1 + I2= 2 A + 2 A= 4 A

Therefore, the total current in the parallel RL circuit is 4 A.

In a parallel RL circuit, the voltage across the resistor and the voltage across the inductor are the same.

However, the current through the resistor and the current through the inductor are not the same, since the current through the inductor lags behind the voltage.

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Related Questions

MIPS
Write a recursive function that compute the Comb(n,r) where n >=r and n,r>=0
Comb(n,r) = 1 if n=r or r=0
Comb(n,r)= Comb(n-1,r) + Comb( n-1,r-1)

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Here is the recursive function that compute the combination (n,r) where n >=r and n,r>=0:```
def Comb(n, r): if r == 0 or n == r: return 1 return Comb(n-1, r) + Comb(n-1, r-1)

In the given code, we have defined a function named 'Comb' that takes two parameters, n and r. Inside the function, we have used an if-else statement to check if r is equal to 0 or n is equal to r. If any of these conditions is true, then the function returns 1. Otherwise, the function calculates the combination of (n,r) using the formula "Comb(n,r)= Comb(n-1,r) + Comb( n-1,r-1)".

In detail, the function works in the following way:1. If r = 0 or n = r, then return 1.2. Otherwise, call the Comb function recursively twice with updated parameters: (n-1, r) and (n-1, r-1).3. Add the results obtained in step 2 and return it. Example: Let's say we want to calculate the combination of (4,2). We will call the 'Comb' function with arguments 4 and 2.

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write a for loop that prints usernum ... -1 0. ex: usernum = -3 outputs:

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This code prompts the user to input a number and stores it in the `usernum` variable. Then, the for loop starts from `usernum` and goes down to -2 (exclusive) with a step size of -1.

The `end=" "` parameter in the `print()` function is used to print the output in a single line separated by spaces.First, we define the starting point of our range using the `usernum` variable that we get from the user input. We then define the end point of our range as -2 (which is exclusive), so the loop will stop at -1. Finally, we set the step size to -1 so that the loop counts down.

The `range()` function returns a sequence of numbers from `usernum` to -2 (exclusive) with a step size of -1. This sequence is used as the input for the for loop.

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Consider the following sequence of memory access where each address is a byte address: 0, 1, 4, 3, 4, 15, 2, 15, 2, 10, 12, 2. Assume that the cash is direct-mapped, cash size is 4 bytes, and block size is two bytes; Map addresses to cache blocks and indicate whether hit or miss.

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The mapping of addresses to cache blocks and the corresponding hit or miss is as follows:

Cache block 0: miss, miss, miss, miss, hit, miss, hit, missCache block 1: miss, miss, miss, miss, miss, hit, miss, hit, miss, miss, miss, hit

Given sequence of memory access where each address is a byte address:

0, 1, 4, 3, 4, 15, 2, 15, 2, 10, 12, 2

Assuming that the cache is direct-mapped, cache size is 4 bytes, and block size is two bytes;

Let us first calculate the number of blocks in the cache.

`Number of blocks in the cache = cache size / block size = 4/2 = 2`

The memory access addresses are as follows:0, 1, 4, 3, 4, 15, 2, 15, 2, 10, 12, 2

The block containing 0 is mapped to the first block (set 0).

This is a cache miss because the first block is empty.

The block containing 1 is mapped to the first block (set 0).

This is a cache miss because the first block contains the block containing 0.

The block containing 4 is mapped to the second block (set 1).

This is a cache miss because the second block is empty.

The block containing 3 is mapped to the second block (set 1).

This is a cache miss because the second block contains the block containing 4.

The block containing 4 is mapped to the second block (set 1).

This is a cache hit because the second block contains the block containing 4.

The block containing 15 is mapped to the first block (set 0).

This is a cache miss because the first block contains the block containing 0.

The block containing 2 is mapped to the second block (set 1).

This is a cache miss because the second block contains the block containing 4.

The block containing 15 is mapped to the first block (set 0).

This is a cache hit because the first block contains the block containing 15.

The block containing 2 is mapped to the second block (set 1).

This is a cache hit because the second block contains the block containing 2.

The block containing 10 is mapped to the first block (set 0).

This is a cache miss because the first block contains the block containing 0.

The block containing 12 is mapped to the second block (set 1).

This is a cache miss because the second block contains the block containing 2.

The block containing 2 is mapped to the second block (set 1).

This is a cache hit because the second block contains the block containing 2.

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Determine the internal normal force, shear force, and tending moment at point C. Assume the reactions at the supports A and B are vertical. 1.5 kN/m 0,5 kN/m B 6 m

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Given: 1.5 kN/m0.5 kN/m6 m In order to determine the internal normal force, shear force, and bending moment at point C, we will determine the reactions at support A and B.

Using the condition of static equilibrium for the vertical direction,Fy = 0RA + RB - 1.5 × 6 - 0.5 × 6 = 0RA + RB = 6 kN …..(1)Now taking moments about point A,MA = 0RA × 6 - 1.5 × 6 × (6/2) - 0.5 × 6 × (6/3) = 0RA = 2.5 kN ……(2)RB = 6 - 2.5 = 3.5 kN ……(3)Calculation of Internal Forces and Bending Moment at point C:

For point C, taking forces to the left as positive and downward forces as positive. FBD of the section CB:

Let us consider a small length dx of section CB at a distance x from support C.

The free body diagram of the section is shown below: Resolving the forces along x and y directions , Fx = 0Nc - F(x) = 0F(x) = Nc …..(4)Fy = 0Vc - 1.5 × x - 0.5 × x + V(x) = 0V(x) = 2x …..(5)Taking moments about point C,MC = 0-M(x) + 1.5 × x × (x/2) + 0.5 × x × (x/3) = 0M(x) = (1/2) × (2/3)x³ - (3/4)x³M(x) = -(1/12)x³ …..(6)The internal normal force is given by : N(x) = - Nc = - (2/3)x³ ……(7)The internal shear force is given by: V(x) = 2x - 1.5x - 0.5x = 0.0N ……(8)The internal bending moment is given by: M(x) = -(1/12)x³ …….(9)Therefore, at point C, Internal normal force, N(x) = - (2/3)x³Internal shear force, V(x) = 0.0     NInternal bending moment, M(x) = -(1/12)x³, where x is the distance measured from support C.

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