Analyse the given electric circuit and answer the network: i) The equivalent resistance of the network T ii) The total electric current I in the circuit

Here's the analysis of the given electric circuit:

i) The equivalent resistance of the network:

Step 1: Identify the resistors and their connection. The circuit shows two resistors, $R_1 = 60 \Omega$ and $R_2 = 30 \Omega$, connected in parallel.

Step 2: Calculate the equivalent resistance ($R_{eq}$) for parallel resistors. The formula for two resistors in parallel is: $R_{eq} = \frac{R_1 R_2}{R_1 + R_2}$

Step 3: Substitute the values and calculate the equivalent resistance. $R_{eq} = \frac{60 \Omega \times 30 \Omega}{60 \Omega + 30 \Omega}$ $R_{eq} = \frac{1800 \Omega^2}{90 \Omega}$ $R_{eq} = 20 \Omega$

Step 4: Select the correct option. Comparing with the given options: A) $30 \Omega$, B) $60 \Omega$, C) $20 \Omega$, D) $90 \Omega$. The correct option is $\boxed{C) 20 \Omega}$.

ii) The total electric current I in the circuit:

Step 1: Identify the total voltage and equivalent resistance. The dry cell provides a voltage $V = 3 \text{ V}$. The equivalent resistance of the circuit is $R_{eq} = 20 \Omega$ (calculated in part i).

Step 2: Apply Ohm's Law to find the total current. Ohm's Law states $I = \frac{V}{R_{eq}}$.

Step 3: Substitute the values and calculate the total current. $I = \frac{3 V}{20 \Omega}$ $I = 0.15 \text{ A}$

Step 4: Select the correct option. Comparing with the given options: A) $0.1 \text{ A}$, B) $0.05 \text{ A}$, C) $0.15 \text{ A}$, D) $0.00 \text{ A}$. The correct option is $\boxed{C) 0.15 A}$.

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