91影视

Ohm鈥檚 law states that the voltage across a conductor is directly proportional to the current flowing through it, provided all physical conditions and temperatures remain constant

$V=IR$

According to Kirchhoff鈥檚 Lawthe sum of the voltages around the closed loop is equal to null.

$\mathbf{鈭�}\mathit{V}\mathbf{=}\mathbf{0}$

Use Ohm鈥檚 law to get resistance value of resistors A and B

$R=\frac{V}{I}$

We get

Resistor A does not obey Ohm鈥檚 law because ohm鈥檚 law states that the resistance remains constant which is not the case here and the V-I graph is not a straight line

Resistor B Obey Ohm鈥檚 law because Ohm鈥檚 law states that the resistance remains constant which is the case here and the V-I graph is a straight line

Power dissipated is given byor$P=VI\mathrm{or}P=\frac{V^2}{R}$

Where Vis the voltage drop across and is the current through the resistor and R is the resistance of the resistor.

For resistor A,

It is connected to battery of$V=40V$ and the current$I=2.35A$

$$\begin{gathered} P=\left(2.35A\right)\left(4.00V\right) \\ P=9.40W \end{gathered}$$

For resistor B,

It is connected to battery of $V=40V$and resistance is $R=3.88惟$

$$\begin{gathered} P=\frac{{\left(4.0V\right)}^2}{3.88惟} \\ P=4.12W \end{gathered}$$

Power dissipated arePower dissipated on the battery A is$P_A=9.40W$

Power dissipated on the battery B is$P_B=4.12W$