A 2V battery is connected across AB as shown in the figure. The value of the current supplied by the battery when in one case, battery’s positive terminal is connected to A and in other case when positive terminal of battery is connected to B will respectively be
A. 0.2 A and 0.1 A
B. 0.1 A and 0.2 A
C. 0.2 A and 0.4 A
D. 0.4 A and 0.2 A
Answer
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Hint: When the positive terminal of battery is connected to A diode ${D}_{1}$ gets forward biased and current flows through it. To find the current flowing through it use Ohm's law. Substitute the values and find the current flowing through the diode ${D}_{1}$. Similarly, when a positive terminal is connected to B, diode ${D}_{2}$ gets forward biased current flows through it. Again, use the formula for Ohm's law and find the current flowing through ${D}_{2}$.
Formula used: $V=IR$
Complete step by step solution:
Given: Voltage (V)= 2V
Resistance ${R}_{1}= 5 \Omega$
Resistance ${R}_{2}= 10 \Omega$
Case.1) When the positive terminal of the battery is connected to A, the diode ${D}_{1}$ gets forward biased while the diode ${D}_{2}$ gets reverse biased. Hence, the current will pass only through diode ${D}_{1}$.
According to Ohm's law,
$V=IR$
Current passing through diode ${D}_{1}$ is given by,
$V= I {R}_{1}$
Substituting values in above equation we get,
$2= I \times 5$
$\Rightarrow I= \dfrac {2}{5}$
$\Rightarrow I= 0.4 A$
Thus, 0.4 current will flow through the diode ${D}_{1}$ when the positive terminal of the battery is connected to A.
Case.2) When the positive terminal of the battery is connected to B, the diode ${D}_{2 }$ gets forward biased while the diode ${D}_{1}$ gets reverse biased. Hence, the current will pass only through diode ${D}_{2}$.
According to Ohm's law,
$V=IR$
Current passing through diode ${D}_{2}$ is given by,
$V= I {R}_{2}$
Substituting values in above equation we get,
$2= I \times 10$
$\Rightarrow I= \dfrac {2}{10}$
$\Rightarrow I= 0.2 A$
Thus, 0.2 current will flow through the diode ${D}_{2 }$ when the positive terminal of the battery is connected to B.
So, the correct answer is “Option D”.
Note: To answer these types of questions, students must know how the current flows through a diode when a positive or negative terminal of a battery is connected to it. They should also know the concept of forward bias and reverse bias. Students generally forget the anode and cathode side of a diode so they should remember that as well. In most of the diodes, current flows only when a positive voltage is given to the anode.
Formula used: $V=IR$
Complete step by step solution:
Given: Voltage (V)= 2V
Resistance ${R}_{1}= 5 \Omega$
Resistance ${R}_{2}= 10 \Omega$
Case.1) When the positive terminal of the battery is connected to A, the diode ${D}_{1}$ gets forward biased while the diode ${D}_{2}$ gets reverse biased. Hence, the current will pass only through diode ${D}_{1}$.
According to Ohm's law,
$V=IR$
Current passing through diode ${D}_{1}$ is given by,
$V= I {R}_{1}$
Substituting values in above equation we get,
$2= I \times 5$
$\Rightarrow I= \dfrac {2}{5}$
$\Rightarrow I= 0.4 A$
Thus, 0.4 current will flow through the diode ${D}_{1}$ when the positive terminal of the battery is connected to A.
Case.2) When the positive terminal of the battery is connected to B, the diode ${D}_{2 }$ gets forward biased while the diode ${D}_{1}$ gets reverse biased. Hence, the current will pass only through diode ${D}_{2}$.
According to Ohm's law,
$V=IR$
Current passing through diode ${D}_{2}$ is given by,
$V= I {R}_{2}$
Substituting values in above equation we get,
$2= I \times 10$
$\Rightarrow I= \dfrac {2}{10}$
$\Rightarrow I= 0.2 A$
Thus, 0.2 current will flow through the diode ${D}_{2 }$ when the positive terminal of the battery is connected to B.
So, the correct answer is “Option D”.
Note: To answer these types of questions, students must know how the current flows through a diode when a positive or negative terminal of a battery is connected to it. They should also know the concept of forward bias and reverse bias. Students generally forget the anode and cathode side of a diode so they should remember that as well. In most of the diodes, current flows only when a positive voltage is given to the anode.
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