The voltage at $B$ in the figure is :

5.3 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5.7 V

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

6.3 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

6 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $5.7 V$
The voltage drop across a Germanium diode is $0.3 V$ .
Hence the potential at point B is $6 V - 0.3 V = 5.7 V$
The current in the circuit would be $\frac{5.7 V - 0 V}{2 k Ω} = 2.85 m A$

Suggest Corrections

Similar questions

Find th evalues of $r_{p} μ$ annd $g_{m}$ of a triode operating at plate voltage 200 $V^{p}$ and grid voltage -6V. The plate characteristics are shown in figure (41-E1).

Q. In a series CR circuit shown in figure, the applied voltage is 10 V and the voltage across capacitor is found to be 8V. Then the voltage across R, and the phase difference between current and the applied voltage will respectively be

Q. In the given circuit, peak voltage across the resistor and capacitor are 6V and 8V. At an instant when voltage across capacitor is 4V in magnitude,

Q. The circuit shown in the figure contains two diodes each with a forward resistance of

50 Ω

with infinite backwards resistance. If the battery voltage is 6V, the current through the

100 Ω

(in amperes) is

A student has to connect 4 cells of 1.5 V each, to form a battery of voltage 6 V. The correct way of connecting these cells is shown in figure.

Join BYJU'S Learning Program

The voltage at B in the figure is :

The correct option is C 5.7VThe voltage drop across a Germanium diode is 0.3V.Hence the potential at point B is 6V−0.3V=5.7VThe current in the circuit would be 5.7V−0V2kΩ=2.85mA

The voltage at $B$ in the figure is :

The correct option is C $5.7 V$
The voltage drop across a Germanium diode is $0.3 V$ .
Hence the potential at point B is $6 V - 0.3 V = 5.7 V$
The current in the circuit would be $\frac{5.7 V - 0 V}{2 k Ω} = 2.85 m A$