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.
