It's Photoelectric Effect, I just a test with this same question. I am not good for explaining exactly how, but I was right.
Answer:
0.4 ohms.
Explanation:
From the circuit,
The voltage reading in the voltmeter = voltage drop across each of the parallel resistance.
1/R' = 1/R1+1/R2
R' = (R1×R2)/(R1+R2)
R' = (2.4×1.2)/(2.4+1.2)
R' = 2.88/3.6
R' = 0.8 ohms.
Hence the current flowing through the circuit is
I = V'/R'................ Equation 1
Where V' = voltmeter reading
I = 6/0.8
I = 7.5 A
This is the same current that flows through the variable resistor.
Voltage drop across the variable resistor = 9-6 = 3 V
Therefore, the resistance of the variable resistor = 3/7.5
Resistance = 0.4 ohms.
So here is my answer. Given that the electric field in <span>a 3.0mm×3.0mm square aluminum wire is 1.0×10−2 V/m, this is how we find the current in the wire.
</span><span>First, we take the distance as 3mm or 0.003m. Using the formula E=V/d, where d is distance, v voltage and E electric field strength, we make V the subject, being V=Ed or 2.2*10^-2*0.003=6.6*10^-5V
</span>Hope this answers your question.
