Solutions: B
Based on individual calculations, the highest effective resistance will only be achieved if the junctions are connected at Q and S.
(NOT connected diagonally at Q and S, which some students thought initially, as it will be short circuit and the resistors are redundant)
Important concepts which you have to know total effective resistance of parallel circuit. Refer to the image below.
Solutions: B
Based on individual calculations, the highest effective resistance will only be achieved if the junctions are connected at Q and S.
(NOT connected diagonally at Q and S, which some students thought initially, as it will be short circuit and the resistors are redundant)
Important concepts which you have to know total effective resistance of parallel circuit. Refer to the image below.
Solution:A
As resistance of the rheostat increases, the total resistance in parallel will increase. Hence potential difference across the parallel bulb Q increases, therefore BRIGHTER.
Total effective resistance of the circuit (include resistance of bulb P) increases, main current thought circuit (bulb P) decreases. Hence bulb P DIMMER.
You can think in terms of p.d of bulb P. Since p.d. across parallel branches increases, p.d across bulb P decreases, hence bulb P is DIMMER.
Without calculation, using resistance and p.d. is easier. To convince yourself, refer to the following diagrams with imaginary numbers.
An a.c. circuit is shown below.
Which of the following diagrams show the voltage V across the resistor?
Solutions: Option C
The power supply is a alternating supply, hence direction of current changes many times in 1 second. (alternating current = a.c.)
You may ask which direction of current should you first assume in order to represent the first ‘hump’?
That is not that important. Rather you must be able to know that the ‘humps’ should alternate as the V across alternates due to the diodes.
When current flows in anticlockwise, current will flow through the 6Ω and the 12Ω (in series). When the current flows in clockwise, current will flow through the 12Ω and the 12Ω (in series). Hence V across the 6Ω and the 12Ω will alternate. Hence B and D are out.
D is out as there is an continuous alternating current flowing through the circuit. Not rectification due to the diodes in the circuit.
The diodes in the 2 branches only stop current from flowing in either branch at any one time, current still continue to flow through the circuit.
Consider when current flows anticlockwise, the V is across the 6Ω. As 6Ω is smaller than the 12Ω in series, the p.d. across 6Ω is smaller. Hence a smaller ‘hump’.
On the other hand, when current flows clockwise, the V is across the 12Ω. As this 12Ω is equal to the 12Ω in series, the p.d. across will be higher than across 6Ω previously. Hence bigger ‘hump’. So C is the answer.
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