Question

Show you work. All answers should be given with two decimal places. 1. A current of 3.33 mA (IMA -0.001A) passes through a resistor which is at a temperature 22.3 °C. As a result, a voltage of 3.00 Volts is generated across the resistor. Calculate the resistance of the resistor in k 2. Rrom 2. Now the same resistor as in question 1 is immersed in liquid nitrogen (T--195.8 °C). The voltage across the resistor is still 3.00V but the current through the resistor is now 2.67 mA Calculate the resistance of the resistor at this new temperature in k 2. Run 3. When a voltage of 4.25 Volts is applied across a light bulb, a current of 122.00 mA passes through its filament. Calculate the resistance, R, of the bulb in ohms (12) and the power, P, in Watts being dissipated in it. R= P= 4. A 3.20 k2 resistor has a current of -1.25 mA passing through it. What is the voltage applied to the resistor? V=

Answer 1

Multiple questions posted, so answering only the first question:

1.According to ohm's law, V=i*R where V is the potential difference across a resistance R and current through it is i

So,R=V/i

In the given problem, V= 3 Volts, I=3.33 mA= 3.33*10^-3 A

So,resistance= 3/(3.33*10^-3)=900.90 ohms.

Answer 2

SOLUTION :

1.

R = V / I = 3.00 / (3.33 * 10^(-3)) = 900.90 Ω = 0.9009 kΩ (ANSWER) 

2.

R = V / I = 3.00 / (2.67 * 10^(-3)) = 1123.60 Ω = 1.1236 kΩ (ANSWER) 

3.

R = V / I = 4.25 / (122 * 10^(-3)) = 34.84 Ω  (ANSWER) 

P = V I = 4,25 * 122*10^(-3) = 0.5185 = 0.52 W (ANSWER).

4.

V = I R = 1.25 * 01^(-3) * 3.20 * 10^3 = 4 volts (ANSWER).