New answers tagged resistors
1
Hey you're getting wrong, the equivalent emf(voltage) of system will be 16V-8V because both have opposite poles facing each other, so their will be net flor of current (-ve to +ve) according to the cell of greater emf(16V cell).Then your total resistance is $$5+1.6+1.4 = 8 \Omega$$(all are in series) . $$I(Current) = E(e.m.f or Voltage)/R(Resistance) = 8/8 = ...
1
Let the resistance of the original wire be R.
R = ρ (L/A)
Now l = L/5
R’ = ρ (L/5A)
Or R’ = R/5
Now 5 resistors of R’ are connected in parallel
1/R(net) = 5/R + 5/R + 5/R + 5/R + 5/R
or
1/2 = 25/R
or
R = 50 Ω.
-2
The wires have resistance. You had 5 resistors in series. If connected in parallel, they give a known parallel resistance.
1
Resistor is anything that pose resistance to flow of charges in a circuit.
Here is how it looks like:
$$R=\rho \dfrac lA$$
Where $\rho$ is material property, $l$ is length and $A$ is cross sectional area.
Here is how 5 resistors in parallel look like:
and the circuit diagram showing the same:
In parallel configuration voltage drop across each ...
2
Seems like you're not making the connection between the actual physical setup and the equations. So, here's a translation:
1) A length of wire is a resistor. By resistor what we mean is that when we apply a potential difference $V$ between the two ends (like from a battery) the resulting current $I$ is given by $V = IR$ where $R$ is the resistance.
2) The ...
1
$\Gamma_{ii}$ is the $i$-th entry on the diagonal of $\Gamma = L^+ = (D-A)^+$, $\Gamma_{jj}$ is the $j$-th entry, and $\Gamma_{ij}$ is the entry located at row $i$, column $j$. Thus $\Omega_{ij}$ is a scalar, but you could assemble all such values into a matrix $\Omega$ that gives the resistances between all pairs of vertices.
Top 50 recent answers are included
