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Physics · 2026 · 3 marks
Semiconductor Electronics: Materials, Devices and Simple CircuitsApplication of Junction Diode as a Rectifier3 marksApplyshort answer
CBSE 2026 · Region 1 · Set 1 · Q25
An ac voltage V_i =12 sin (100 π t ) V is applied between points A and B in a network of two ideal diodes and three resistors as shown in figure.…
An ac voltage $\displaystyle \mathrm{V}_{\mathrm{i}}=12 \sin (100 \pi \mathrm{t}) \mathrm{V}$ is applied between points A and B in a network of two ideal diodes and three resistors as shown in figure. During the positive half-cycle of the input voltage $\displaystyle \mathrm{V}_{\mathrm{i}}$ supplied to the network.
(a)
Identify which of the two diodes will conduct and why ?
(b)
Redraw an equivalent circuit diagram to show the flow of current.
(c)
Calculate the output voltage drops $\displaystyle \mathrm{V}_{0}$ across the three resistors when the input voltage attains its peak value.
Marking-scheme solution
(a)
Diode $\displaystyle D_2$ is in forward bias and it conducts.
(b)
(c)
Voltage drop across RP, PB and RB is
$\displaystyle V_{RP} = 4\,\mathrm{V}$ , $\displaystyle V_{PB} = 8\,\mathrm{V}$, $\displaystyle V_{RB} = 12\,\mathrm{V}$
Alternatively:
If the student considers positive AC cycle reaching diode $\displaystyle D_1$
(a)
Diode $\displaystyle D_1$ is in forward bias and it conducts.
(b)
(c)
$\displaystyle V_{PB} = 12\,\mathrm{V},\ V_{PR} = 3\,\mathrm{V},\ V_{RB} = 9\,\mathrm{V}$
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CBSE Class 12 Physics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.