Question

The circuit fragment shown in the figure below is called a voltage divider. (a) If R_load is not attached, show that V_out = VR_2/(R_1 + R_2). (Do this on paper. Your instructor may ask you to turn in this work.) (b) If R_1 = R_2 = 9.5 k Ohm, what is the smallest value of R_load that can be used so V_out drops by less than 11 percent from its unloaded value? (V_out is measured with respect to ground.) k Ohm

Answer 1

DATA:

$R_{1}=R_{2}=9.5\,k\Omega$

SOLUTION

Applying KVL at the left side we have:

Simillarly, applying KVL at the right side we have:

Isolating from equation (1) :

$I=\frac{V}{R_{1}+R_{2}}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (3)$

Replacing (3) into equation (2) we obtain:

${\color{Blue} V_{out}=\frac{VR_{2}}{R_{1}+R_{2}}}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (4)$

Part (b)

After attaching the Load Resistance $R_{load}$ at the circuit, we have that resistor $R_{2}$ are connected in parallel with $R_{load}$ , so the equivalent resistance $R_{eq}$ is:

$R_{eq}=R_{1}+\frac{R_{2}R_{load}}{R_{2}+R_{load}}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (5)$

Now, the current throughout the circuit is:

$I=\frac{V}{R_{eq}}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (6)$

Replacing (5) into (6) we have:

$I=\frac{V(R_{2}+R_{load})}{R_{1}R_{2}+R_{1}R_{load}+R_{2}R_{load}}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (7)$

Due to $R_{2}$ and $R_{load}$ are connected in parallel, the voltage is the same, so:

$I_{2}R_{2}=I_{load}R_{load}$

Isolating $I_{2}$ we get:

$I_{2}=\frac{I_{load}R_{load}}{R_{2}}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (8)$

Now, applying KCL at the circuit we have:

Replacing (7) and (8) into equation (9) we get:

$\frac{V(R_{2}+R_{load})}{R_{1}R_{2}+R_{1}R_{load}+R_{2}R_{load}}=\frac{I_{load}R_{load}}{R_{2}}+I_{load }\, \, \, \, \, \, \, \,\, \, \, \, \, \, \, \, \, (10)$

Isolating $I_{load}$ we get:

$I_{load}=\frac{VR_{2}}{R_{1}R_{2}+R_{1}R_{load}+R_{2}R_{load}}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (11)$

Now, the voltage $V'_{out}$ across the load resistor is:

Replacing (11) into equation (12) we get:

$V'_{out}=\frac{VR_{2}R_{load}}{R_{1}R_{2}+R_{1}R_{load}+R_{2}R_{load}}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (13)$

But;

Replacing (14) into (13) we have:

$0.89V_{out}=\frac{VR_{2}R_{load}}{R_{1}R_{2}+R_{1}R_{load}+R_{2}R_{load}}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (15)$

Replacing (4) into (15) we get:

$\frac{0.89VR_{2}}{R_{1}+R_{2}}=\frac{VR_{2}R_{load}}{R_{1}R_{2}+R_{1}R_{load}+R_{2}R_{load}}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (16)$

Now, isolating $R_{load}$ we obtain:

$R_{load}=\frac{0.91R_{1}R_{2}}{0.09(R_{1}+R_{2})}$

finally, replacing data given values:

$R_{load}=\frac{0.91(9.5\,k\Omega)(9.5\,k\Omega)}{0.09(9.5\,k\Omega+9.5\,k\Omega)}$

Solving we obtain:

${\color{Blue} R_{load}=48.03\,k\Omega=48\,k\Omega}$