The given circuit has two ideal diodes connected as shown in the figure below. The current flowing through the resistance $R_{1}$ will be

Question

The given circuit has two ideal diodes connected as shown in the figure below- The current flowing through the resistance R-1 will be2-5 A10-0 A3-13 A1-43 A

Toppr Admin · Accepted Answer

From figure- ideal diode D1 is reversed biased whereas ideal-xA0-diode D2 is forward biased- Thus-xA0-D1 acts as an open switch while-xA0-D2 as a closed switch as shown in the equivalent circuit-Thus current flowing through R1- -xA0- -xA0- -xA0- -xA0- -xA0-I-VR1-R3-102-2-2-5A-xA0-

The given circuit has two ideal diodes connected as shown in the figure below. The current flowing through the resistance $R_{1}$ will be

$2.5 A$
$10.0 A$
$3.13 A$
$1.43 A$

From figure, ideal diode $D_{1}$ is reversed biased whereas ideal diode $D_{2}$ is forward biased. Thus $D_{1}$ acts as an open switch while $D_{2}$ as a closed switch as shown in the equivalent circuit.
Thus current flowing through $R_{1}$ , $I = \frac{V}{R_{1} + R_{3}} = \frac{10}{2 + 2} = 2.5 A$

The given circuit has two ideal diodes connected as shown in the figure below. The current flowing through the resistance R1 will be2.5A10.0A3.13A1.43A

From figure, ideal diode D1 is reversed biased whereas ideal diode D2 is forward biased. Thus D1 acts as an open switch while D2 as a closed switch as shown in the equivalent circuit.Thus current flowing through R1, I=VR1+R3=102+2=2.5A

The given circuit has two ideal diodes connected as shown in the figure below. The current flowing through the resistance $R_{1}$ will be

$2.5 A$
$10.0 A$
$3.13 A$
$1.43 A$

From figure, ideal diode $D_{1}$ is reversed biased whereas ideal diode $D_{2}$ is forward biased. Thus $D_{1}$ acts as an open switch while $D_{2}$ as a closed switch as shown in the equivalent circuit.
Thus current flowing through $R_{1}$ , $I = \frac{V}{R_{1} + R_{3}} = \frac{10}{2 + 2} = 2.5 A$