The value of V1 is ____.

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Understand the Problem

The question is asking for the value of V1 in an electrical circuit based on given resistances and voltages. To solve it, we will need to analyze the circuit using Ohm's Law and potentially Kirchhoff's voltage law.

Answer

$V_1 \approx 7.98 \, V$
Answer for screen readers

The value of ( V_1 ) is approximately ( 7.98 , V ).

Steps to Solve

  1. Identify Given Values
    We need to identify the voltages and resistances in the circuit.
  • Voltage sources: ( 27 , V ) and ( 5 , V )
  • Voltage drop: ( 9 , V )
  • Resistor values: ( 2.2 , k\Omega ), ( 0.56 , k\Omega ), and ( 1.2 , k\Omega )
  1. Apply Kirchhoff's Voltage Law
    According to Kirchhoff's Voltage Law, the sum of all voltages in a closed loop must equal zero. The loop can be expressed as: $$ 27 , V - V_1 - 9 , V - \left( I \cdot 2.2 \times 10^3 \Omega \right) - 5 , V = 0 $$

  2. Calculate Total Resistance and Current
    We need to calculate the total resistance in the circuit using: $$ R_{total} = 2.2 , k\Omega + 0.56 , k\Omega + 1.2 , k\Omega = 4.96 , k\Omega $$
    Then, calculate the current ( I ) using Ohm's Law: $$ I = \frac{V_{total}}{R_{total}} = \frac{27 , V - 5 , V}{4.96 , k\Omega} $$

  3. Calculate Voltage Drop across ( 2.2 , k\Omega )
    The voltage drop across the ( 2.2 , k\Omega ) resistor can be calculated as: $$ V_{drop} = I \cdot R = I \cdot 2.2 , k\Omega $$

  4. Find ( V_1 )
    Substituting the current value back into the equation we formed in step 2 allows us to find ( V_1 ): $$ V_1 = 27 , V - 9 , V - V_{drop} - 5 , V $$

The value of ( V_1 ) is approximately ( 7.98 , V ).

More Information

This calculation considers how voltage, resistance, and current interact in a circuit. Understanding these relationships is crucial for analyzing complex circuits.

Tips

  • Forgetting to convert ( k\Omega ) to ( \Omega ) in calculations: Ensure all resistances are in the same units (i.e., ( \Omega )).
  • Incorrectly applying Kirchhoff's Law: Remember to account for the signs of each voltage in the loop properly.

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