Hello. This may be a simple problem, but I'm struggling to obtain the resistance R found in the figure below.
First what's throwing my are the two voltage sources. I'm used to seeing these present in op-amps, but the way they intersect is throwing me. I'm not sure how to determine their cummulative affect.
My first impulse was to find the voltage drop across the 6K resistor, but since there is no current value given I can't do this.
I then considered nodal analysis, denoting the red node as Vb.
\[ \frac{12-V_b}{6K} + \frac{7-V_b}{0} = \frac{V_b}{R_{eq}} \]
However, since the 7V has no resistance, can I just eliminate it from the equation? Therefore:
\[ \frac{12-V_b}{6K} = \frac{V_b}{R_{eq}} \]
If this is correct, then I can solve for R by calculating the equivalent resistance.
\[ R_{eq} = \frac{10kOhm R}{10kOhm+R} \]
I am mainly suspicious that I'm overcomplicating this and that the nodal analysis is incorrect/unnecessary.
Thanks in advance.
First what's throwing my are the two voltage sources. I'm used to seeing these present in op-amps, but the way they intersect is throwing me. I'm not sure how to determine their cummulative affect.
My first impulse was to find the voltage drop across the 6K resistor, but since there is no current value given I can't do this.
I then considered nodal analysis, denoting the red node as Vb.
\[ \frac{12-V_b}{6K} + \frac{7-V_b}{0} = \frac{V_b}{R_{eq}} \]
However, since the 7V has no resistance, can I just eliminate it from the equation? Therefore:
\[ \frac{12-V_b}{6K} = \frac{V_b}{R_{eq}} \]
If this is correct, then I can solve for R by calculating the equivalent resistance.
\[ R_{eq} = \frac{10kOhm R}{10kOhm+R} \]
I am mainly suspicious that I'm overcomplicating this and that the nodal analysis is incorrect/unnecessary.
Thanks in advance.