To answer the questions regarding the circuit with given parameters, let's go through each question step by step.
2.1 Finding the value of resistor R:
You mentioned that the total resistance of the circuit is 4.8 Ω and there seems to be a resistance of 4R involved. Without loss of generality, let's assume that the resistors are in series. The total resistance \( R_{total} \) can be represented as:
\[ R_{total} = R + 4R = 5R \]
Given that \( R_{total} = 4.8 , \Omega \): \[ 5R = 4.8 , \Omega \]
To find the value of \( R \), we can solve for \( R \): \[ R = \frac{4.8 , \Omega}{5} = 0.96 , \Omega \]
Thus:
- Answer 1: 0.96 Ω
2.2 Reading on the voltmeter:
You mentioned that the current through the 4R resistor is 1.8 A. Assuming that the resistor is purely resistive, the voltage across it (V) can be calculated using Ohm's law, \( V = I \times R \).
Let's express \( 4R \): \[ 4R = 4 \times 0.96 , \Omega = 3.84 , \Omega \]
Now calculate the voltage: \[ V = I \times (4R) = 1.8 , A \times 3.84 , \Omega = 6.912 , V \]
So if rounding to a whole number:
- Answer 2: 7 V (if rounding)
2.3 Energy converted in the resistor 4R in 2 minutes:
The power \( P \) dissipated in the resistor can be calculated using the formula: \[ P = I^2 \times R \]
Substituting our values: \[ P = (1.8 , A)^2 \times 3.84 , \Omega = 3.24 \times 3.84 = 12.4416 , W \]
To find the energy \( E \) converted in two minutes: \[ E = P \times t \]
Where \( t \) is the time in seconds. 2 minutes is 120 seconds. \[ E = 12.4416 , W \times 120 , s = 1493.992 , J \approx 1494 , J \]
Thus:
- Answer 3: 1494 J
In summary:
- Answer 1: 0.96 Ω
- Answer 2: 7 V
- Answer 3: 1494 J
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