A wire of resistance 5.9 Ω is connected to a battery whose emf ε is 4.0 V and whose internal resistance is 1.2 Ω. In 2.9 min, ho
1 answer:
You might be interested in
Answer:
A) the ammeter is x
B)
- voltage across R₁ (left resistor) = 0.75 V
- voltage across the right one = 0.3 V
C) 1.05 V
Explanation:
From the diagram attached below;
A) Assuming the homes were wired in series, and one of the homes face short circuit then all the houses would face power cut but it doesn't happen. So they must be connected in parallel.
Therefore; The ammeter is connected in series, Hence, the ammeter is x and the voltmeter must be z.
B)
Given that:
x = 0.15 A
z = 0.3 V
Resistor (R) on the left = 5 ohms
Then, voltage across R₁ (left resistor) = 5×(x)
= 5×0.15
= 0.75 V
voltage across the right one = z = 0.3 V
C)
The total voltage of battery = 0.75+0.3 = 1.05 V
