Question

In electrical circuit theory, a circuit C(t) is called a linear circuit if it satisfies the superposition principle given by C(at1+ bt2 ) =aC(t1 )+ bC(t2 ) , where a,b are
constants. Show that the circuit C(t) = 3t is linear.

Answer 1

we have to show that the circuit C(t) = 3t is linear .

according electrical theory, a circuit C(t) is called a linear circuit if it satisfies the superposition principle given by, $C(at_1+bt_2)=aC(t_1)+bC(t_2)$ where a and b are constant and $t_1,t_2\in\textbf{domain of C(t)}$

Let's take two point $t_1$ and $t_2$ from domain of C(t).

now, $C(at_1+bt_2)=3(at_1+bt_2)$

$aC(t_1)=3at_1$

and $bC(t_2)=3bt_2$

here it is clear that,

$3(at_1+bt_2)=3at_1+3bt_2$

or, $C(at_1+bt_2)=aC(t_1)+bC(t_2)$

hence, C(t) is linear .

Answer 2

Answer:-

We have to show that the circuit

C(t) = 3t is linear .

According electrical theory:

Circuit C(t) is called a linear circuit if it satisfies the superposition principle given by,

C(at1+bt2) = aC(t1) + bC(t2)

where "a" and "b" are constant and

t1, t2 € domain of C(t).

Let's take two points t1 and t2 from domain of C(t).

Now,

C (at1+bt2) = 3 (at1+bt2)

aC (t1) = 3at1

and, bC (t2)= 3bt2

Here it is clear that,

3 (at1+bt2) = 3at1 + 3bt2

Or,

C (at1+bt2) = aC(t1) + bC(t2)

Hence,

C(t) is linear.