QUESTION 2 A battery of emf 24 V and negligible internal resistance is connected to a resistor network as shown in the circuit diagram in the diagram below. (a) Show that the resistance of the single equivalent resistor that could replace the four resistors between the points A and B is 50Omega (b) If R is 50Omega calculate (i) The current in R.

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QUESTION 2 A battery of emf 24 V and negligible internal resistance is connected to a resistor network as shown in the circuit diagram in the diagram below. (a) Show that the resistance of the single equivalent resistor that could replace the four resistors between the points A and B is 50Omega (b) If R is 50Omega calculate (i) The current in R.

Answer 1

### (a) The resistance of the single equivalent resistor is $50\Omega$. ### (b)(i) The current in $R$ is $0.48\text{ A}$.

Answer 2

## Step 1: Identify the Resistor Network Configuration ### The resistor network consists of four resistors between points A and B. We need to determine if they are in series or parallel and then find the equivalent resistance. ## Step 2: Calculate the Equivalent Resistance ### Given that the equivalent resistance between points A and B is $50\Omega$, we can assume that the resistors are arranged in such a way that their combined resistance is $50\Omega$. This can be verified by analyzing the circuit diagram, which is not provided here, but we accept the given equivalent resistance. ## Step 3: Apply Ohm's Law to Find the Current ### Ohm's Law states that $I = \frac{V}{R}$. Given the emf (voltage) of the battery is 24 V and the equivalent resistance $R$ is $50\Omega$, we can calculate the current. \[I = \frac{24\text{ V}}{50\Omega} = 0.48\text{ A}\]

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