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In the given circuit, what is the voltage across the load resistance [tex]R_L[/tex] if the resistance values are 1.5 kilo ohms and 2.5 kilo ohms?

Answer :

The voltage across the load resistance [tex]\(R_L\)[/tex] is 5.45 volts, calculated by applying Ohm's law with the total current and load resistance values.

Given:

Resistance [tex]\(R_1 = 1.5 \, k\Omega\)[/tex]

Resistance [tex]\(R_2 = 2.5 \, k\Omega\)[/tex]

The total resistance in the circuit [tex](\(R_{\text{total}}\))[/tex] can be calculated using the formula for resistors in series:

[tex]\[ R_{\text{total}} = R_1 + R_2 \][/tex]

Substituting the given values:

[tex]\[ R_{\text{total}} = 1.5 \, k\Omega + 2.5 \, k\Omega = 4 \, k\Omega \][/tex]

The total current [tex](\(I_{\text{total}}\))[/tex] flowing in the circuit can be calculated using Ohm's law:

[tex]\[ I_{\text{total}} = \frac{V_{\text{total}}}{R_{\text{total}}} \][/tex]

We know that the voltage across the load resistance [tex]\(R_L\)[/tex] is the same as the total voltage [tex](\(V_{\text{total}}\))[/tex], therefore:

[tex]\[ V_{\text{total}} = V_{R_L} \][/tex]

Substituting the given voltage source value:

[tex]\[ 10V = V_{R_L} \][/tex]

Substituting the calculated total resistance [tex](\(R_{\text{total}}\))[/tex] into Ohm's law equation:

[tex]\[ I_{\text{total}} = \frac{10V}{4 \, k\Omega} = 2.5 \, mA \][/tex]

Now, we can use Ohm's law to find the voltage across the load resistance [tex](\(V_{R_L}\))[/tex]:

[tex]\[ V_{R_L} = I_{\text{total}} \times R_L \][/tex]

Substituting the calculated total current [tex](\(I_{\text{total}}\))[/tex] and the load resistance

([tex]R_L[/tex]):

[tex]\[ V_{R_L} = 2.5 \, mA \times 2.18 \, k\Omega = 5.45 \, volts \][/tex]

Therefore, the voltage across the load resistance [tex]\(R_L\)[/tex] is 5.45 volts.

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