PYQs / GATE EE / 2017 / Set 1 / Q14 GATE EE 2017 Set 1 — Question 14 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. MCQ +1 / -0.33 Medium Rectifiers (HW, FW, Bridge) Diode Circuits Analog & Digital Electronics
Analog & Digital Electronics → Diode Circuits → Rectifiers (HW, FW, Bridge)
Last updated 5 September 2026
Question For the circuit shown in the figure below, assume that diodes
D 1 D_1 D 1 ,
D 2 D_2 D 2 and
D 3 D_3 D 3 are ideal.
The DC components of voltages
v 1 v_1 v 1 and
v 2 v_2 v 2 , respectively are
Correct answer (B) -0.5 V and 0.5 V
Solution 1. The parallel combination of ideal diodes D 1 D_1 D 1 D 2 D_2 D 2 v ( t ) v(t) v ( t ) 2. For the positive half-cycle (v ( t ) > 0 v(t) > 0 v ( t ) > 0 D 3 D_3 D 3 The circuit is a voltage divider with two equal resistors R R R v 1 ( t ) = v ( t ) ⋅ R R + R = v ( t ) 2 v_1(t) = v(t) \cdot \frac{R}{R+R} = \frac{v(t)}{2} v 1 ( t ) = v ( t ) ⋅ R + R R = 2 v ( t ) v 2 ( t ) = v ( t ) ⋅ R R + R = v ( t ) 2 v_2(t) = v(t) \cdot \frac{R}{R+R} = \frac{v(t)}{2} v 2 ( t ) = v ( t ) ⋅ R + R R = 2 v ( t ) 3. For the negative half-cycle (
v ( t ) < 0 v(t) < 0 v ( t ) < 0 ):
D 3 D_3 D 3 v 1 ( t ) = v ( t ) ⋅ R R + 0 = v ( t ) v_1(t) = v(t) \cdot \frac{R}{R+0} = v(t) v 1 ( t ) = v ( t ) ⋅ R + 0 R = v ( t ) v 2 ( t ) = 0 v_2(t) = 0 v 2 ( t ) = 0 D 3 D_3 D 3 4. DC component of
v 1 v_1 v 1 :
V 1 , D C = 1 T ∫ 0 T v 1 ( t ) d t = 1 T [ ∫ 0 T / 2 π sin ( ω t ) 2 d t + ∫ T / 2 T π sin ( ω t ) d t ] V_{1,DC} = \frac{1}{T} \int_0^T v_1(t) dt = \frac{1}{T} \left[ \int_0^{T/2} \frac{\pi \sin(\omega t)}{2} dt + \int_{T/2}^T \pi \sin(\omega t) dt \right] V 1 , D C = T 1 ∫ 0 T v 1 ( t ) d t = T 1 [ ∫ 0 T /2 2 π s i n ( ω t ) d t + ∫ T /2 T π sin ( ω t ) d t ] Using
∫ 0 T / 2 sin ( ω t ) d t = T π \int_0^{T/2} \sin(\omega t) dt = \frac{T}{\pi} ∫ 0 T /2 sin ( ω t ) d t = π T and
∫ T / 2 T sin ( ω t ) d t = − T π \int_{T/2}^T \sin(\omega t) dt = -\frac{T}{\pi} ∫ T /2 T sin ( ω t ) d t = − π T :
V 1 , D C = 1 T [ π 2 ⋅ T π + π ⋅ ( − T π ) ] = 0.5 − 1 = − 0.5 V V_{1,DC} = \frac{1}{T} \left[ \frac{\pi}{2} \cdot \frac{T}{\pi} + \pi \cdot \left(-\frac{T}{\pi}\right) \right] = 0.5 - 1 = -0.5\text{ V} V 1 , D C = T 1 [ 2 π ⋅ π T + π ⋅ ( − π T ) ] = 0.5 − 1 = − 0.5 V .
5. DC component of v 2 v_2 v 2 V 2 , D C = 1 T ∫ 0 T v 2 ( t ) d t = 1 T [ ∫ 0 T / 2 π sin ( ω t ) 2 d t + 0 ] = 1 T [ π 2 ⋅ T π ] = 0.5 V V_{2,DC} = \frac{1}{T} \int_0^T v_2(t) dt = \frac{1}{T} \left[ \int_0^{T/2} \frac{\pi \sin(\omega t)}{2} dt + 0 \right] = \frac{1}{T} \left[ \frac{\pi}{2} \cdot \frac{T}{\pi} \right] = 0.5\text{ V} V 2 , D C = T 1 ∫ 0 T v 2 ( t ) d t = T 1 [ ∫ 0 T /2 2 π s i n ( ω t ) d t + 0 ] = T 1 [ 2 π ⋅ π T ] = 0.5 V .
Thus, the DC components are
− 0.5 V -0.5\text{ V} − 0.5 V and
0.5 V 0.5\text{ V} 0.5 V .
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Correct answer (B) -0.5 V and 0.5 V
Solution 1. The parallel combination of ideal diodes D 1 D_1 D 1 D 2 D_2 D 2 v ( t ) v(t) v ( t ) 2. For the positive half-cycle (v ( t ) > 0 v(t) > 0 v ( t ) > 0 D 3 D_3 D 3 The circuit is a voltage divider with two equal resistors R R R v 1 ( t ) = v ( t ) ⋅ R R + R = v ( t ) 2 v_1(t) = v(t) \cdot \frac{R}{R+R} = \frac{v(t)}{2} v 1 ( t ) = v ( t ) ⋅ R + R R = 2 v ( t ) v 2 ( t ) = v ( t ) ⋅ R R + R = v ( t ) 2 v_2(t) = v(t) \cdot \frac{R}{R+R} = \frac{v(t)}{2} v 2 ( t ) = v ( t ) ⋅ R + R R = 2 v ( t ) 3. For the negative half-cycle (
v ( t ) < 0 v(t) < 0 v ( t ) < 0 ):
D 3 D_3 D 3 v 1 ( t ) = v ( t ) ⋅ R R + 0 = v ( t ) v_1(t) = v(t) \cdot \frac{R}{R+0} = v(t) v 1 ( t ) = v ( t ) ⋅ R + 0 R = v ( t ) v 2 ( t ) = 0 v_2(t) = 0 v 2 ( t ) = 0 D 3 D_3 D 3 4. DC component of
v 1 v_1 v 1 :
V 1 , D C = 1 T ∫ 0 T v 1 ( t ) d t = 1 T [ ∫ 0 T / 2 π sin ( ω t ) 2 d t + ∫ T / 2 T π sin ( ω t ) d t ] V_{1,DC} = \frac{1}{T} \int_0^T v_1(t) dt = \frac{1}{T} \left[ \int_0^{T/2} \frac{\pi \sin(\omega t)}{2} dt + \int_{T/2}^T \pi \sin(\omega t) dt \right] V 1 , D C = T 1 ∫ 0 T v 1 ( t ) d t = T 1 [ ∫ 0 T /2 2 π s i n ( ω t ) d t + ∫ T /2 T π sin ( ω t ) d t ] Using
∫ 0 T / 2 sin ( ω t ) d t = T π \int_0^{T/2} \sin(\omega t) dt = \frac{T}{\pi} ∫ 0 T /2 sin ( ω t ) d t = π T and
∫ T / 2 T sin ( ω t ) d t = − T π \int_{T/2}^T \sin(\omega t) dt = -\frac{T}{\pi} ∫ T /2 T sin ( ω t ) d t = − π T :
V 1 , D C = 1 T [ π 2 ⋅ T π + π ⋅ ( − T π ) ] = 0.5 − 1 = − 0.5 V V_{1,DC} = \frac{1}{T} \left[ \frac{\pi}{2} \cdot \frac{T}{\pi} + \pi \cdot \left(-\frac{T}{\pi}\right) \right] = 0.5 - 1 = -0.5\text{ V} V 1 , D C = T 1 [ 2 π ⋅ π T + π ⋅ ( − π T ) ] = 0.5 − 1 = − 0.5 V .
5. DC component of v 2 v_2 v 2 V 2 , D C = 1 T ∫ 0 T v 2 ( t ) d t = 1 T [ ∫ 0 T / 2 π sin ( ω t ) 2 d t + 0 ] = 1 T [ π 2 ⋅ T π ] = 0.5 V V_{2,DC} = \frac{1}{T} \int_0^T v_2(t) dt = \frac{1}{T} \left[ \int_0^{T/2} \frac{\pi \sin(\omega t)}{2} dt + 0 \right] = \frac{1}{T} \left[ \frac{\pi}{2} \cdot \frac{T}{\pi} \right] = 0.5\text{ V} V 2 , D C = T 1 ∫ 0 T v 2 ( t ) d t = T 1 [ ∫ 0 T /2 2 π s i n ( ω t ) d t + 0 ] = T 1 [ 2 π ⋅ π T ] = 0.5 V .
Thus, the DC components are
− 0.5 V -0.5\text{ V} − 0.5 V and
0.5 V 0.5\text{ V} 0.5 V .
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