Let's analyze the circuit and the options step by step: ## **Given:** - Voltage source, \( V = 3V \) - \( R_1 = 2\Omega \) (First branch) - \( R_2 = 2\Omega \) (Second branch) - Both resistors are in **parallel** with the voltage source. --- ### **Step 1: Understand Parallel Circuits** In a parallel circuit: - The voltage across each resistor is **equal** to the source voltage. So, the voltage across \( R_1 \) is \( 3V \). --- ### **Step 2: Calculate Current in the First Branch (\( I_1 \))** Using Ohm's Law: \[ I_1 = \frac{V}{R_1} \] Plug in the values: \[ I_1 = \frac{3V}{2\Omega} = 1.5A \] --- ### **Step 3: Examine Each Option** #### **Option 1:** > I1 = V/R1 = 1A - **Calculation:** \( \frac{3V}{2\Omega} = 1.5A \), so this is **incorrect**. - **Explanation:** The calculation doesn't match the correct value. #### **Option 2:** > I1 = V/R1 = 1.5A - **Calculation:** \( \frac{3V}{2\Omega} = 1.5A \), so this is **correct**. - **Explanation:** This matches the calculation using Ohm's Law. #### **Option 3:** > I1 = V/R1 = 3A - **Calculation:** \( \frac{3V}{2\Omega} = 1.5A \), not 3A, so this is **incorrect**. - **Explanation:** The calculation is not correct according to Ohm's Law. #### **Option 4:** > No answer text provided. - **Explanation:** Not applicable. --- ## **Final Answer** \[ \boxed{\text{Option 2 is correct: } I_1 = 1.5A} \] ### **Why Other Options are Incorrect:** - **Option 1:** Incorrect calculation (\( 1A \) instead of \( 1.5A \)). - **Option 3:** Incorrect calculation (\( 3A \) instead of \( 1.5A \)). - **Option 4:** No answer provided. --- **Summary Table:** | Option | Value | Correct? | Reason | |--------|-------|----------|-----------------------------------------------------| | 1 | 1A | ❌ | Incorrect calculation | | 2 | 1.5A | ✅ | Correct (matches Ohm's Law) | | 3 | 3A | ❌ | Incorrect calculation | | 4 | -- | ❌ | No answer text provided |

# Step-by-Step Solution for the Circuit Problem --- ## **Given Data:** - Voltage source: \( V_s = 12\,\text{V} \) - Resistor \( R_1 = 12\,\Omega \) - Resistor \( R_2 = 24\,\Omega \) *The resistors are connected in parallel across the voltage source.* --- ## **Step 1: Understand the Circuit** In a parallel circuit: - The voltage across each resistor is **equal** to the source voltage. Thus: \[ V_{R_1} = V_{R_2} = V_s = 12\,\text{V} \] --- ## **Step 2: Calculate Current through \( R_1 \) (\( I_1 \))** Using **Ohm's Law**: \[ I_1 = \frac{V_{R_1}}{R_1} \] Plugging in the values: \[ I_1 = \frac{12\,\text{V}}{12\,\Omega} = 1.0\,\text{A} \] --- ## **Step 3: Verify with \( R_2 \) and Total Current** - Current through \( R_2 \): \[ I_2 = \frac{12\,\text{V}}{24\,\Omega} = 0.5\,\text{A} \] - Total current supplied by the source: \[ I_{total} = I_1 + I_2 = 1.0\,\text{A} + 0.5\,\text{A} = 1.5\,\text{A} \] - Confirm with equivalent resistance: \[ R_{eq} = \left( \frac{1}{R_1} + \frac{1}{R_2} \right)^{-1} = \left( \frac{1}{12} + \frac{1}{24} \right)^{-1} = \left( \frac{2 + 1}{24} \right)^{-1} = \left( \frac{3}{24} \right)^{-1} = 8\,\Omega \] - Check total current: \[ I_s = \frac{V_s}{R_{eq}} = \frac{12\,\text{V}}{8\,\Omega} = 1.5\,\text{A} \] which matches the sum of branch currents. --- ## **Step 4: Final Answer** The **current through the first resistor \( R_1 \)** is: \[ \boxed{ I_1 = 1.0\,\text{A} } \] --- ## **Why the Other Options Are Incorrect:** | Option | Value | Explanation | |---------|---------|--------------| | **1. \( I_1 = 1\,\text{A} \)** | **Correct** | Calculated directly using Ohm's Law. | | **2. \( I_1 = 1.5\,\text{A} \)** | Incorrect | This is the *total* current supplied by the source, not the current through \( R_1 \). | | **3. \( I_1 = 3\,\text{A} \)** | Incorrect | Would require \( R_1 = 12\,\Omega \) with \( V=36\,\text{V} \), which is not the case. | | **4. No answer provided** | Incorrect | The proper answer is 1.0 A, so this choice is invalid. | --- ## **Summary:** - **Current through \( R_1 \):** **1 A** - **Total source current:** 1.5 A - **Method used:** Ohm's Law in parallel circuit. --- **Note:** Always verify resistor values and source voltage to accurately compute branch currents.