Question

in a circuit a 12V battery with internal resistance of 2ohm is connected to three resistors with resistances of 10.6, 4 and 7 ohms. resistors are serial. two diodes one Si and the other Ge connected parallel to each other. This set of diodes is connected serially to the above circuit between the 7 and 4 ohm resistors. Vbe of Si is 0.4V and Vbe of Ge is 0.2V. Find the electric current which flows through the circuit and the voltages across each diode.

Answer 1

Answer on Question #65455, Physics / Electric Circuits

$$I = \frac {E 1}{R _ {\text {t o t a l}} + R _ {\text {i n t e m a l}}}$$ $$R _ {\text {t o t a l}} = R 1 + R 2 + R _ {V D 1 V D 2} + R 3 = 2 1. 6 + R _ {V D 1 V D 2}$$

Since the diodes are in parallel and have different forward voltage, the diode with smaller forward voltage appears first and all the current will flow through it follows from this that the voltage drop on the diode Assembly is equal to $0.2\mathrm{V}$ (forward voltage of a germanium diode) and so the resistance of the diode assemblies is equal to the resistance of open a germanium diode, it follows:

$$R _ {V D 1 V D 2} = R _ {V D 2} = \frac {0 . 2}{I}$$ $$I = \frac {E 1}{R _ {\text {t o t a l}} + R _ {\text {i n t e m a l}}} = \frac {1 2}{2 1 . 6 + \frac {0 . 2}{I} + 2} = \frac {1 2}{2 3 . 6 + \frac {0 . 2}{I}}$$ $$I = \frac {1 2}{2 3 . 6 + \frac {0 . 2}{I}}$$ $$I \cdot (2 3. 6 + \frac {0 . 2}{I}) = 1 2$$ $$2 3. 6 \cdot I + 0. 2 = 1 2$$ $$2 3. 6 \cdot I = 1 1. 8$$ $$I = 0. 5$$

answer: $I = 0.5A$ , $V_{VD1} = V_{VD1} = 0.2V$