# IGCSE PHYSICS ELECTRIC CIRCUITS QUANTITIES

```ELECTRIC CIRCUITS QUANTITIES
1. VOLTAGE V (V):
POTENTIAL DIFFERENCE IN A PARALLEL CIRCUIT:
IF YOU WEREN’T GIVEN THE EMF OF THE BATTERY, THEN USE THIS
FORMULA: V = TOTAL I × TOTAL R
IF YOU WERE GIVEN THE EMF OF THE BATTERY, THEN THE VOLTAGE IS
SIMPLY THE SAME AS THE EMF OF THE BATTERY, GIVEN THAT THE
COMPONENT IS CONNECTED IN PARALLEL WITH THE BATTERY! (MAKE
SURE YOU LOOK AT THE CIRCUIT VERY CAREFULLY, AS THIS CAN BE VERY
TRICKY). IF IT’S NOT CONNECTED TO THE BATTERY DIRECTLY, THEN YOU
WILL SIMPLY USE THE FORMULA: p.d. = SPECIFIC I × SPECIFIC R
SO FOR EXAMPLE:
In this circuit, the p.d. across the 3Ω
resistor is 12V, as the 3Ω resistor is
connected to the battery directly,
meaning it’s in parallel with the battery,
therefore, it has the same p.d. across it
as the EMF of the battery which is
12V.
In this circuit however, the p.d. across
the 2Ω resistor is NOT 6V like the EMF
of the battery, as the 2Ω resistor is not
connected to the battery directly (as
there is a resistor on the right hand
side), meaning it’s in series with the
battery, and therefore, the p.d. across
it is NOT the same as the EMF of the
battery.
POTENTIAL DIFFERENCE IN A SERIES CIRCUIT:
IF YOU WEREN’T GIVEN THE EMF OF THE BATTERY / IF YOU WERE
GIVEN THE EMF OF THE BATTERY AND YOUR COMPONENT IS NOT
CONNECTED DIRECTLY TO YOUR BATTERY, i.e. CONNECTED IN
SERIES WITH THE BATTERY, USE THIS FORMULA: V = TOTAL /
SPECIFIC I (AS YOU CAN HAVE A PARALLEL CIRCUIT IN A SERIES
CIRCUIT) × SPECIFIC R (RESISTANCE OF THIS CERTAIN COMPONENT
THAT YOU ARE TRYING TO FIND THE p.d. ACROSS IT)
If you were asked to find the p.d. between
the terminals X and Y in this circuit, the p.d.
in the strands of wire above and below
terminals X and Y would be the same as it’s
a parallel circuit (JUST THE BIT BETWEEN
TERMINALS X AND Y). To get the p.d. in
the series circuit which is the strand of wire
above terminals X and Y, you will have to
simply multiply the total current (that was
given in the question as the reading of the
ammeter) by the specific resistance 3Ω. This
p.d. will be same in the series circuit that’s
below terminals X and Y, as the two series
circuits above and below terminals X and Y
are in parallel.
IF YOU WERE GIVEN THE EMF OF THE BATTERY, THEN SIMPLY, THE EMF
OF THE BATTERY IS SPLIT BETWEEN THE COMPONENTS, WITH THE ONE
THAT HAS THE HIGHEST Ω WILL GET THE HIGHEST V. TO CALCULATE
HOW MUCH EACH COMPONENT RECEIVES, USE RATIOS!!! BE AWARE OF
THIS POINT: IF THERE WERE FOR EXAMPLE 2 RESISTORS, YOU CAN GET
THE p.d. ACROSS ONE OF THEM AND THEN SUBTRACT THIS p.d. FROM
THE TOTAL EMF OF THE BATTERY, TO GET THE p.d. ACROSS THE OTHER
ONE.
2. RESISTANCE R (Ω):
TOTAL / EFFECTIVE RESISTANCE OF RESISTORS IN SERIES:
RT or Rp = R1 + R2 + R3 + R4 + ……………………………….
TOTAL / EFFECTIVE RESISTANCE OF RESISTORS IN PARALLEL: (TOTAL RESISTANCE
WILL LESS THAN THE RESISTANCE OF THE LEAST RESISTOR):
2 RESISTORS IN PARALLEL:
RT or Rp = (R1×R2) / (R1+R2)
3 OR MORE RESISTORS IN PARALLEL:
RT or Rp = RECIPROCAL OF (1/R1 + 1/R2 + 1/R3 + ………..)
IF YOU ADD MORE RESISTORS IN PARALLEL, THE TOTAL Ω DECREASES.
IN A PARALLEL CIRCUIT, IF YOU ADD IDENTICAL RESISTORS IN PARALLEL, THEY
WILL HAVE THE SAME CURRENT AND SAME VOLTAGE ACROSS THEM AS THEY
HAVE THE SAME Ω.
Be careful in this circuit
while finding the combined
resistance, R, of the
resistors. Firstly, find the R
of the resistors which are
together. That would be
the total R of the resistors
in parallel, the 1Ω and the
total R of these resistors to
the resistor in series, which
is the 2Ω resistor.
RESISTOR(S) IN A SERIES CIRCUIT:
Voltage (leftover) / Total Current
RESISTOR(S) IN A PARALLEL CIRCUIT:
Total EMF of the battery / Specific (TOTAL) current flowing in
through it (them) in this strand of wire
3.
CURRENT I (A):
CURRENT IN A SERIES CIRCUIT:
I = TOTAL EMF / TOTAL R
CURRENT IN A PARALLEL CIRCUIT/ IF COMPONENT IS CONNECTED
TO THE BATTERY DIRECTLY:
TOTAL CURRENT (I1) = SUM OF ALL SPECIFIC CURRENTS OR
EMF / TOTAL Ω
SPECIFIC CURRENT = P.D (USUALLY EMF, HOWEVER, IT WILL
NOT BE EMF IF THE COMPONENT IS NOT CONNECTED
DIRECTLY TO THE BATTERY, FOR EXAMPLE IF THERE IS A
RESISTOR ON THE SIDE, IN WHICH CASE YOU WILL HAVE TO
USE THE FORMULA SPECIFIC CURRENT × VOLTAGE LEFT OVER
FROM THE VOLTAGE TAKEN BY THE RESISTOR WHICH WILL BE
THE SAME ACROSS THE OTHER PARALLEL STRANDS OF WIRE) /
SPECIFIC R (RESISTANCE OF THIS CERTAIN COMPONENT (S)
THAT YOU ARE TRYING TO FIND THE CURRENT FLOWING
THROUGH IT / THEM)
MAKE SURE YOU KNOW THE RELATIONSHIPS BETWEEN CURRENTS IN AN
ELECTRIC CIRCUIT
For example in this circuit,
relationships would be:
I1 = 14
I2 + I3 = I4
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