2 edited body edited Mar 26 '17 at 21:30 freecharly 13.9k22 gold badges88 silver badges3131 bronze badges The easiest way to solve this problem ist to use Thevenin's theorem for the open terminals which arise when you cut open the circuit where the resistance $$R_3$$ is inserted. It is easily seen that the Thevenin voltage will be the battery voltage $$V_B$$ $$V_{th}= V_B$$ and the Thevenin resistance is the parallel circuit resistance of $$R_1$$ and $$R_2$$ $$R_{th}^{-1}=R_{1}^{-1}+ R_{1}^{-1}$$$$R_{th}^{-1}=R_{1}^{-1}+ R_{2}^{-1}$$ Thus the resistance of the circuit as seen from the battery is $$R=R_{th}+R_3$$ and the current flowing through the battery is $$I_B=\frac{V_B}{R}$$ The easiest way to solve this problem ist to use Thevenin's theorem for the open terminals which arise when you cut open the circuit where the resistance $$R_3$$ is inserted. It is easily seen that the Thevenin voltage will be the battery voltage $$V_B$$ $$V_{th}= V_B$$ and the Thevenin resistance is the parallel circuit resistance of $$R_1$$ and $$R_2$$ $$R_{th}^{-1}=R_{1}^{-1}+ R_{1}^{-1}$$ Thus the resistance of the circuit as seen from the battery is $$R=R_{th}+R_3$$ and the current flowing through the battery is $$I_B=\frac{V_B}{R}$$ The easiest way to solve this problem ist to use Thevenin's theorem for the open terminals which arise when you cut open the circuit where the resistance $$R_3$$ is inserted. It is easily seen that the Thevenin voltage will be the battery voltage $$V_B$$ $$V_{th}= V_B$$ and the Thevenin resistance is the parallel circuit resistance of $$R_1$$ and $$R_2$$ $$R_{th}^{-1}=R_{1}^{-1}+ R_{2}^{-1}$$ Thus the resistance of the circuit as seen from the battery is $$R=R_{th}+R_3$$ and the current flowing through the battery is $$I_B=\frac{V_B}{R}$$ 1 answered Mar 26 '17 at 21:25 freecharly 13.9k22 gold badges88 silver badges3131 bronze badges The easiest way to solve this problem ist to use Thevenin's theorem for the open terminals which arise when you cut open the circuit where the resistance $$R_3$$ is inserted. It is easily seen that the Thevenin voltage will be the battery voltage $$V_B$$ $$V_{th}= V_B$$ and the Thevenin resistance is the parallel circuit resistance of $$R_1$$ and $$R_2$$ $$R_{th}^{-1}=R_{1}^{-1}+ R_{1}^{-1}$$ Thus the resistance of the circuit as seen from the battery is $$R=R_{th}+R_3$$ and the current flowing through the battery is $$I_B=\frac{V_B}{R}$$