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We know that: A voltage divider is a simple circuit which turns a large voltage into a smaller one. Using just two series resistors and an input voltage, we can create an output voltage that is a fraction of the input.

But I am unable to figure out \$V_x\$ across the 6 ohm resistor.

The 12 and 6 ohm resistors in parallel threw me off.

rc circuit

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    \$\begingroup\$ \$ \Large V_X = V_C * \frac{12\Omega || 6\Omega }{(12\Omega || 6\Omega ) + 8\Omega} = V_C * \frac{\frac{12\Omega}{3}}{ \frac{12\Omega}{3} + 8\Omega } = V_C * \frac{4\Omega }{4 + 8\Omega} = V_C * \frac{4\Omega}{12\Omega} \$ \$\endgroup\$ – G36 Jun 20 '19 at 6:23
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    \$\begingroup\$ What you should do is calculate the equivalent resistance of 12 ohms and 6 ohms in parallel, which happens to be 4 ohms, and substitute that in the schematic. G36's comment here shows how to do this. \$\endgroup\$ – Bart Jun 20 '19 at 6:50
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    \$\begingroup\$ Just curious. Why was the value for the capacitor provided, unless you have an exponential decay taking place that you need to address in the answer? \$\endgroup\$ – jonk Jun 20 '19 at 6:53
  • \$\begingroup\$ Thanks y'all I think I understand now \$\endgroup\$ – Mattizzy Jun 20 '19 at 13:58
  • \$\begingroup\$ Yes I am calculating the exponential decay @jonk \$\endgroup\$ – Mattizzy Jun 22 '19 at 1:13
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G36 is correct. I'll try to provide a little explanation to what he's saying.

Since the 12 ohm and 6 ohm resistors are in parallel, the voltage across both of them is the same \$V_x\$ Volts. So you can find out the equivalent resistance as \$R_{eq} = 12||6 = \frac{12*6}{12+6} = 4\$. So now, you have a 4 ohm resistor in series with a 8 ohm resistor. Using voltage division, you get \$V_x = V_c * \frac{4}{4+8} = \frac{V_c}{3}\$.

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  • \$\begingroup\$ What if the 12 ohms resistor were to be in series, do you still need to find the resultant of 12 ohms and 6 ohms resistor? \$\endgroup\$ – Mattizzy Jun 22 '19 at 1:12
  • \$\begingroup\$ In that case, you would take the equivalent of the 12 and 8 ohm resistors since the voltage drop across them would not be the same. Since the two are in parallel here, the voltage drop across them is the same and we were able to consider them as a single equivalent resistor. You cannot do that if they are in parallel. \$\endgroup\$ – Prateek Dhanuka Jun 23 '19 at 17:05
  • \$\begingroup\$ I don't want to be a little off but I was thinking that since 6 ohm and 12 ohm are in series I was thinking we can just apply voltage division right away like. V_x = V_c * \frac{6}{6+12} = \frac{V_c}{3} \$\endgroup\$ – Mattizzy Jun 25 '19 at 2:29
  • \$\begingroup\$ I don't want to be a little off but I was thinking that since 6 ohm and 12 ohm are in series I was thinking we can just apply voltage division right away like\$V_x = V_c * \frac{6}{6+12} = \frac{V_c}{3}\$ \$\endgroup\$ – Mattizzy Jun 25 '19 at 2:37
  • \$\begingroup\$ But then what about the 8 ohm resistor. Since even that is connected in series it has an affect. The circuit can be considered as a 8 ohm is series with 12 ohm = 20 ohm resistor. After this you can use voltage division and you get \$V_x = V_c * \frac{6}{6+20} = \frac{3*V_c}{13}\$ \$\endgroup\$ – Prateek Dhanuka Jun 25 '19 at 4:37

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