Question

You are given a semiconductor resistor made from silicon with an impurity concentration of resistivity 1.00×10−3ωm. the resistor has a height of h =0.5 mm, a length of l = 2 mm, and a width of w = 1.25 mm. the resistor can absorb (dissipate) up to p = 7.81w. what is the resistance of the resistor (r), the maximum voltage (v), and the maximum current (i)?

Answer 1

The resistance of the semiconductor resistor is 16 Ω. The maximum voltage across the resistor is 7.808 V. The maximum current through the resistor is 0.488 A.

Step-by-step explanation:

To calculate the resistance of the semiconductor resistor, we can use the formula R = ρL/A, where ρ is the resistivity of the silicon, L is the length of the resistor, and A is the cross-sectional area. Plugging in the given values, R = (1.00 × 10-3 Ωm)(0.002 m) / (0.5 × 10-3 m)(1.25 × 10-3 m) = 16 Ω.

The maximum voltage across the resistor can be calculated using the equation V = IR, where I is the maximum current. Since the resistor can dissipate up to 7.81 W, and the resistance is 16 Ω, we have 7.81 W = I (16 Ω), which gives us I = 0.488 A. Therefore, V = (0.488 A)(16 Ω) = 7.808 V.

Finally, to find the maximum current, we can use Ohm's law, I = V/R. Using the values obtained above, I = 7.808 V / 16 Ω = 0.488 A.

Answer 2

Answer: Resistance =
`3.2 \Omega` , Current = 1.56 A, Voltage =4.99 V

The resistance,

`R=\frac {\rho l}{A}`

where,
`\rho` is resistivity, A is the area and l is the length of the resistor.

It is given that:

`\rho=1.0*10^(-3)\Omega m`

Length, l=2 mm

Area,
`A= width * height=1.25 mm* 0.5 mm=0.625 mm^2`

Hence,
`R=(1.0*10^(-3)\Omega m * 2*10^(-3)m)/(0.625*10^(-6)m^2)=3.2\Omega`

We know, Power,
`P=I^2R`

`\Rightarrow I=\sqrt{(P)/(R)}`

`P=7.81 W`

`I=\sqrt{\frac {7.81 W}{3.2\Omega}}=√(2.44)A=1.56A`

We know, Voltage,
`V=IR=1.56*3.2=4.99 V`