What will be the required length of a nichrome wire used as a heater coil and having a resistance of 2 Ohm/m for a heater of 1 kW at 200 V? If the current entering the terminal of... What will be the required length of a nichrome wire used as a heater coil and having a resistance of 2 Ohm/m for a heater of 1 kW at 200 V? If the current entering the terminal of an element is given by i = (6t^2 - 2t) A, then the total charge entering the positive terminal between t = 1 s and t = 3 s will be equal to: Read more

Steps to Solve

1. Calculate the required length for the heater coil

To find the required length of the nichrome wire, we use the formula for electrical power:

$$ P = \frac{V^2}{R} $$

Where:

• ( P = 1000 , \text{W} ) (1 kW)
• ( V = 200 , \text{V} )

Rearranging this formula to solve for the resistance ( R ):

$$ R = \frac{V^2}{P} $$

Substituting the values:

$$ R = \frac{(200)^2}{1000} $$

Calculating ( R ):

$$ R = \frac{40000}{1000} = 40 , \Omega $$

Since the resistance per meter of the nichrome wire is given as ( 2 , \Omega/m ), we can use the formula for total resistance:

$$ R = r \cdot L $$

Where:

• ( r = 2 , \Omega/m )
• ( L = ? )

Setting the equations equal:

$$ 40 = 2 \cdot L $$

Solving for ( L ):

$$ L = \frac{40}{2} = 20 , \text{m} $$

2. Calculate the total charge entering the terminal

The charge ( Q ) can be found using the equation:

$$ Q = \int i , dt $$

Given the current equation:

$$ i = 6t^2 - 2t $$

We need to find the charge between ( t = 1 , s ) and ( t = 3 , s ):

$$ Q = \int_{1}^{3} (6t^2 - 2t) , dt $$

Calculating the integral:

First, find the antiderivative:

$$ \int (6t^2 - 2t) , dt = 2t^3 - t^2 + C $$

Now, evaluate this from ( t = 1 ) to ( t = 3 ):

$$ Q = \left[ 2(3^3) - (3^2) \right] - \left[ 2(1^3) - (1^2) \right] $$

Calculating each part:

$$ Q = \left[ 2 \cdot 27 - 9 \right] - \left[ 2 \cdot 1 - 1 \right] $$

$$ Q = \left[ 54 - 9 \right] - \left[ 2 - 1 \right] $$

$$ Q = 45 - 1 = 44 , C $$