Question

How do you calculate the load and bleeder currents in a voltage divider? Vs = 10V. R1 & R2 both 10,000 ohms each.

3. Connect a 100 kS2 load to the circuit. Apply power, and measu age (½). Use la to calculate: . The load and bleeder currents.

Answer 1

Ans) In the above given circuit ,it is given that $R_{1}=R_{2}=10000\: \: \Omega$

and $R_{L}=100\: \: k\Omega$

Using voltage division rule ,the load voltage in the above circuit is given as

$V_{L}=V_{s}*\frac{R_{2}\parallel R_{L}}{R_{1}+R_{2}\parallel R_{L}}$

$V_{L}=V_{s}*\frac{\frac{R_{2}* R_{L}}{R_{2}+ R_{L}}}{R_{1}+\frac{R_{2}* R_{L}}{R_{2}+ R_{L}}}$

$V_{L}=V_{s}*\frac{R_{2}* R_{L}}{R_{1}(R_{2}+ R_{L})+R_{2}* R_{L}}$

Now substitute the values we get

$V_{L}=10*\frac{10000* 100k}{10000*(10000+ 100k)+10000*100k}$

The bleeder current is given as

$I_{Bleeder}=\frac{V_{L}}{R_{2}}=\frac{4.762}{10000}=476.2\: \: \mu A$

$I_{Bleeder}=476.2\: \: \mu A$

Load current is given as

$I_{L}=\frac{V_{L}}{R_{L}}=\frac{4.762}{100k}=47.62\: \: \mu A$

$I_{L}=47.62\: \: \mu A$