17 — D.c. circuits Flashcards
Potentiometer formula
V1 = R1/(R1+R2)xV0
Series
Resistance:
Re= R1 + R2 + …Rn
Current:
I = V/R
Potential difference across 1 resistor:
V = IR
Current: same
Pd: sum
Achieves higher effective resistance than parallel arrangement
Bulb usually less bright than parallel
Parallel circuit resistance
Resistance:
Re = [1/R1 + 1/R2 + … + 1/Rn]^-1
Current:
I = I1 + I2 +… In
Potential difference across any resistor:
Ve
Pd same
Current sum
Bulb usually brighter than n series cus volt identical
Current splits across diff parallel branches -> current thru each cell = V/R -> when more parallel branches added, current is overdrawn from power source -> power depletes quickly
Transducers
Convert changes in physical conditions, such as pressure, tempt or light, into electrical signals to allow the measurement of these physical conditions
Eg thermistors, light-dependent resistors, photovoltaic cells and piezoelectric sensors
Thermistors
Negative temperature coefficient thermistors -> tempt increases, resistance decreases
LDRs
Light intensity increases, resistance decreases
Define potential divider
A potential divider is a voltage divider, which makes use of the voltage drop across resistors in series to divide voltage.
Explain why the pd across lamp is not proportional to the current in it
The lamp is a non-ohmic conductor. The lamp’s resistance increases as the tempt increases, resulting in pd across lamp not proportional to current.
Explain how the student gradually increases the current in the lamp form zero using potentiometer.
By moving the jockey of the potentiometer from Q to P, potential diff across bulb would increase from 0V to 12V thus current increases from 0V.
Explain why the brightness of lamp increases when the light source is placed next to LDR and the thermistor is heated.
- Potentiometer method
As LI increases, resistance of LDR decreases -> pd of LDR decreases. As tempt increases, resistance of thermistor decreases -> pd of thermistor decreases -> pd across bulb increases -> bulb brighter. - Current in a series circuit method
As LI increases, resistance of LDR decreases -> pd of LDR decreases. As tempt increases, resistance of thermistor decreases. Since total effective resistance decreases, by V=IR, pd increases, brightness increases.
Describe and explain how the brightness of bulb changes when env brightness decreases.
Under dim lighting, resistance of a LDR increases and since it is connected in parallel to a fixed resistor, total resistance will increase. From potential divider concept, the pd across the combined setup or bulb will increase which also increase its brightness.
Explain why the voltmeter reading increases as tempt rises
When tempt increases, resistance of thermistor decreases. The pd across thermistor decreases thus pd across the fixed resistor will increase. Therefore, the voltmeter will register an increase in reading.
Parallel circuit. Filament of one of the bulb breaks. Comment on wut happens to the ammeter reading and brightness of the bulb.
Ammeter reading decreases since total effective resistance of circuit increases.
The brightness of the bulb remains the same as the pd of the bulbs in parallel branches remains the same. Assuming resistance remains constant, the current in each branch will also remain the same. Hence, brightness remains the same.
State why the pd across the 2 fixed resistors r the same
Since both resistors have the same resistance of x ohm, the emf is shared equally among the 2 resistors.
State an advantage of using a potential divider circuit than a series circuit to control the brightness of the lamp.
Both circuits can control the brightness of the lamp. However, a potential meter circuit may short-circuit the lamp and therefore switch off the lamp, which a series circuit cannot achieve.
